#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int MAXN = 500005;
// --- Persistent Segment Tree ---
struct Node {
int count;
Node *left, *right;
Node(int count, Node *left, Node *right) : count(count), left(left), right(right) {}
};
Node* null;
Node* roots[MAXN];
// Build an empty tree
Node* build(int l, int r) {
if (l == r) return new Node(0, null, null);
int mid = l + (r - l) / 2;
return new Node(0, build(l, mid), build(mid + 1, r));
}
// Update the tree (Add 1 at position p)
Node* update(Node* node, int l, int r, int p) {
if (l == r) return new Node(node->count + 1, null, null);
int mid = l + (r - l) / 2;
if (p <= mid) return new Node(node->count + 1, update(node->left, l, mid, p), node->right);
else return new Node(node->count + 1, node->left, update(node->right, mid + 1, r, p));
}
// Standard Query: Count in range [ql, qr] for a specific version
int query(Node* node, int l, int r, int ql, int qr) {
if (ql > r || qr < l) return 0;
if (ql <= l && r <= qr) return node->count;
int mid = l + (r - l) / 2;
return query(node->left, l, mid, ql, qr) + query(node->right, mid + 1, r, ql, qr);
}
// --- Critical Helper Function ---
// Find smallest index 'x' (in range of A-values) such that:
// Count of students with A <= x AND B in [yl, yr] >= k
int search_tree(Node* node_l, Node* node_r, int l, int r, int yl, int yr, int k) {
// If total students in this full node range [yl, yr] < k, we can't find such x here
// Note: This logic is slightly different. We are searching for 'x' (time/version),
// but the tree is built on B-values.
// Actually, to search 'x' efficiently, we usually just Binary Search on the versions.
// However, O(log^2 N) is acceptable.
// A strict O(log N) requires a different tree structure (Tree of B values, nodes store A).
// But for IOI Teams, simple Binary Search on the answer 'x' is standard and passes.
// We will implement Binary Search in the 'solve' function for clarity and safety.
return 0;
}
int N;
vector<int> adj[MAXN];
// Wrapper to count students with A <= x AND B in [y1, y2]
int count_students(int x, int y1, int y2) {
if (y1 > y2) return 0;
return query(roots[x], 1, N, y1, y2);
}
// Initial Setup
void init(int n, int A[], int B[]) {
N = n;
null = new Node(0, nullptr, nullptr);
null->left = null->right = null;
roots[0] = build(1, N);
for (int i = 0; i < N; i++) {
adj[A[i]].push_back(B[i]);
}
for (int i = 1; i <= N; i++) {
roots[i] = roots[i-1];
for (int b : adj[i]) {
roots[i] = update(roots[i], 1, N, b);
}
}
}
// DP State
int dp[MAXN]; // Stores the "slack" (available - used)
int K_sorted[MAXN];
// Binary Search to find the split point between index i and j in the stack
// We want to find the coordinate 'x' where the transition from i equals transition from j
int find_split(int i, int j) {
// We want smallest x such that:
// count(x, K[i]+1, K[j]) >= dp[j] - dp[i]
// where K[i] < K[j].
int target = dp[j] - dp[i];
int l = K_sorted[j], r = N + 1; // Range of possible A-values
while (l < r) {
int mid = l + (r - l) / 2;
if (count_students(mid, K_sorted[i] + 1, K_sorted[j]) >= target) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
int can(int M, int K[]) {
// 1. Sort Groups
for(int i=0; i<M; i++) K_sorted[i] = K[i];
sort(K_sorted, K_sorted + M);
// 2. Setup Stack
// Stack stores indices of K_sorted
vector<int> st;
// Use a dummy base case
// We treat index -1 as having K=0, dp=0
// But to avoid negative indices, let's manually handle the first element logic
// or just assume implicit base.
// Let's use clean iteration.
// DP definition: dp[i] = max students remaining after satisfying groups 0...i
// dp[i] = min_{j < i} ( dp[j] + count(K[i], K[j]+1, K[i]) ) - K[i]
// Re-mapped logic to match standard variables:
// We actually just compute values on the fly.
st.clear();
// Dummy element representing K=0, dp=0
// We can't push a real index, so we handle the first step separately
// or insert a dummy into K_sorted?
// Inserting dummy is safer.
vector<int> k_real;
k_real.push_back(0);
for(int i=0; i<M; i++) k_real.push_back(K_sorted[i]);
// Reset DP array (only need size M+1)
// Note: In global scope, dp is size MAXN, but here we index by 0..M
// We'll use a local vector or reuse the global one carefully.
vector<int> local_dp(M + 1, 0);
st.push_back(0); // Index 0 is the dummy (K=0, dp=0)
for (int i = 1; i <= M; i++) {
int k = k_real[i];
// Stack Maintenance
while (st.size() > 1) {
int j = st.back(); // Top
int l = st[st.size() - 2]; // Second Top
// Check if 'j' is redundant.
// It is redundant if the split point between l and j is >= split point between j and i
// Or simpler: find split point between l and j. If that split is > k, j is useless?
// Correct Logic:
// The split point is where the "influence" of l ends and j begins.
// If that split point is to the right of 'k', then for current 'k', l is strictly better/same as j.
int split = find_split(l, j); // Uses global K_sorted logic, need to adapt to k_real
// Careful with variable mapping. Let's fix find_split to take direct K values or indices into k_real.
// Inline Find Split for indices l and j in k_real:
int target = local_dp[j] - local_dp[l];
int low = k_real[j], high = N + 1;
int split_pos = N + 1;
while(low < high) {
int mid = low + (high - low)/2;
if (count_students(mid, k_real[l] + 1, k_real[j]) >= target) {
split_pos = mid;
high = mid;
} else {
low = mid + 1;
}
}
if (split_pos <= k) {
st.pop_back();
} else {
break;
}
}
int j = st.back();
// Calculate DP[i]
// dp[i] = dp[j] + count(k, K[j]+1, k) - k
int valid_cnt = count_students(k, k_real[j] + 1, k);
local_dp[i] = local_dp[j] + valid_cnt - k;
if (local_dp[i] < 0) return 0; // Impossible
st.push_back(i);
}
return 1;
}
