#include <bits/stdc++.h>
using namespace std;
#define int long long
#define F first
#define S second
#define pb push_back
#define endl '\n'
#define all(v) v.begin(),v.end()
typedef vector<int> vi;
typedef tuple<int,int,int> iii;
typedef vector<vi> vvi;
typedef vector<iii> viii;
struct SegmentTree{
int n;
vi tree, lazy;
SegmentTree(int _n){
n = _n;
tree.assign(4*n + 5, 0);
lazy.assign(4*n + 5, 0);
}
void push(int node,int s,int e){
if (lazy[node] == 0) return;
tree[node] += lazy[node] * (e - s + 1);
if (s != e){
lazy[node*2] += lazy[node];
lazy[node*2+1] += lazy[node];
}
lazy[node] = 0;
}
void update(int node,int s,int e,int l,int r,int val){
push(node,s,e);
if (s > e || r < s || e < l) return;
if (l <= s && e <= r){
lazy[node] += val;
push(node,s,e);
return;
}
int m = (s+e)/2;
update(node*2,s,m,l,r,val);
update(node*2+1,m+1,e,l,r,val);
tree[node] = tree[node*2] + tree[node*2+1];
}
int query(int node,int s,int e,int x){
push(node,s,e);
if (s > e || x < s || e < x) return 0;
if (s == e && s == x) return tree[node];
int m = (s+e)/2;
return query(node*2,s,m,x) + query(node*2+1,m+1,e,x);
}
};
int n, m, k;
vi sector;
vi meteors;
viii queries;
vvi devlet;
int32_t main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m;
sector.assign(m, 0);
devlet.assign(n, vi());
for (int i = 0; i < m; ++i){
cin >> sector[i];
sector[i]--;
devlet[ sector[i] ].pb(i);
}
meteors.assign(n, 0);
for (int i = 0; i < n; ++i) cin >> meteors[i];
cin >> k;
queries.resize(k+1);
for (int i = 1; i <= k; ++i){
int l,r,x; cin >> l >> r >> x;
queries[i] = make_tuple(l-1, r-1, x);
}
// parallel binary search boundaries: [1, k+1], answer k+1 means "NIE"
vi left(n, 1), right(n, k+1);
// number of iterations needed = ceil(log2(k+1))
int LOG = 0;
while ((1LL << LOG) <= k) ++LOG;
for (int iter = 0; iter < LOG; ++iter){
SegmentTree tree(m);
vector<vi> check(k+2); // check[t] -> list of states testing mid == t
for (int i = 0; i < n; ++i){
if (left[i] <= right[i] - 1){
int mid = (left[i] + right[i]) / 2;
check[mid].pb(i);
}
}
// apply queries incrementally for times 1..k
for (int t = 1; t <= k; ++t){
int a,b,val; tie(a,b,val) = queries[t];
if (a <= b){
tree.update(1, 0, m-1, a, b, val); // FIX: include a==b (single sector)
} else {
// wrap-around
tree.update(1, 0, m-1, a, m-1, val);
tree.update(1, 0, m-1, 0, b, val);
}
for (int state : check[t]){
long long sum = 0;
for (int sec : devlet[state]){
sum += tree.query(1, 0, m-1, sec);
if (sum >= meteors[state]) break; // small micro-optimization
}
if (sum >= meteors[state]){
right[state] = t;
} else {
left[state] = t + 1;
}
}
}
}
for (int i = 0; i < n; ++i){
if (left[i] == k+1) cout << "NIE\n";
else cout << left[i] << '\n';
}
return 0;
}
