// not my code
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 100000;
const int LOG = 18; // since 2^17 = 131072 > 100000
int up[MAXN+1][LOG];
ll sumCap[MAXN+1][LOG];
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int N, Q;
cin >> N >> Q;
vector<int> D(N+1), C(N+1);
for(int i = 1; i <= N; i++){
cin >> D[i] >> C[i];
}
// parent[i] = index of the next reservoir below i with strictly larger diameter (or 0)
vector<int> parent(N+1, 0);
stack<int> st;
for(int i = N; i >= 1; i--){
while(!st.empty() && D[st.top()] <= D[i]) {
st.pop();
}
parent[i] = st.empty() ? 0 : st.top();
st.push(i);
}
// Binary lifting tables:
// up[i][k] = the 2^k-th ancestor of i along the "next-greater" chain
// sumCap[i][k] = sum of capacities of the 2^k nodes starting at i along that chain
for(int i = 1; i <= N; i++){
up[i][0] = parent[i];
sumCap[i][0] = C[i];
}
// ensure up[0][*] = 0, sumCap[0][*] = 0 by static initialization
for(int k = 1; k < LOG; k++){
for(int i = 1; i <= N; i++){
int p = up[i][k-1];
up[i][k] = (p == 0 ? 0 : up[p][k-1]);
sumCap[i][k] = sumCap[i][k-1]
+ (p == 0 ? 0LL : sumCap[p][k-1]);
}
}
// Answer each query independently
while(Q--){
int R;
ll V;
cin >> R >> V;
ll rem = V;
int curr = R;
// Jump as far as we can while the total capacity of the jumped-over
// reservoirs is still less than the remaining water.
for(int k = LOG - 1; k >= 0; k--){
if(curr != 0 && up[curr][k] != 0 && sumCap[curr][k] < rem){
rem -= sumCap[curr][k];
curr = up[curr][k];
}
}
// Now curr is the highest node we can reach without exceeding rem.
// The next reservoir to fill is curr itself:
int answer = 0;
if(curr != 0 && C[curr] >= rem) {
answer = curr;
} else {
// Either curr==0 (we fell off) or its capacity < rem.
// In both cases, the water flows to waterways.
answer = 0;
}
cout << answer << "\n";
}
return 0;
}
