Let a_1, …, a_n be an array of n positive integers. In one operation, you can choose an index i such that a_i = i, and remove a_i from the array (after the removal, the remaining parts are concatenated).

The weight of a is defined as the maximum number of elements you can remove.

You must answer q independent queries (x, y): after replacing the x first elements of a and the y last elements of a by n+1 (making them impossible to remove), what would be the weight of a?

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 3 ⋅ 10^5) — the length of the array and the number of queries.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — elements of the array.

The i-th of the next q lines contains two integers x and y (x, y ≥ 0 and x+y < n).

Output

Print q lines, i-th line should contain a single integer — the answer to the i-th query.

Examples

Input


13 5
2 2 3 9 5 4 6 5 7 8 3 11 13
3 1
0 0
2 4
5 0
0 12


Output


5
11
6
1
0


Input


5 2
1 4 1 2 4
0 0
1 0


Output


2
0

Note

Explanation of the first query:

After making first x = 3 and last y = 1 elements impossible to remove, a becomes [×, ×, ×, 9, 5, 4, 6, 5, 7, 8, 3, 11, ×] (we represent 14 as × for clarity).

Here is a strategy that removes 5 elements (the element removed is colored in red):

  * [×, ×, ×, 9, \color{red}{5}, 4, 6, 5, 7, 8, 3, 11, ×] 
  * [×, ×, ×, 9, 4, 6, 5, 7, 8, 3, \color{red}{11}, ×] 
  * [×, ×, ×, 9, 4, \color{red}{6}, 5, 7, 8, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 7, \color{red}{8}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, \color{red}{7}, 3, ×] 
  * [×, ×, ×, 9, 4, 5, 3, ×] (final state) 



It is impossible to remove more than 5 elements, hence the weight is 5.
#include <bits/stdc++.h>
using namespace std;
void setIO(string s) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
int N, Q;
int arr[300001];
int BIT[2 * 300001];
vector<pair<int, int> > queries[300001];
int ans[300001];
void update(int x, int v) {
  for (; x <= 2 * N; x += x & -x) {
    BIT[x] += v;
  }
}
int query(int x) {
  int ans = 0;
  for (; x > 0; x -= x & -x) {
    ans += BIT[x];
  }
  return ans;
}
int main() {
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    int x;
    cin >> x;
    arr[i] = i - x;
  }
  for (int i = 1; i <= Q; i++) {
    int x, y;
    cin >> x >> y;
    int r = N - y;
    queries[r].push_back({x + 1, i});
  }
  for (int r = 1; r <= N; r++) {
    if (arr[r] >= 0) {
      int lmin = 0;
      int lmax = r + 1;
      while (lmax - lmin > 1) {
        int mid = (lmin + lmax) / 2;
        if (arr[r] <= query(mid)) {
          lmin = mid;
        } else {
          lmax = mid;
        }
      }
      update(1, 1);
      update(lmin + 1, -1);
    }
    for (auto p : queries[r]) {
      ans[p.second] = query(p.first);
    }
  }
  for (int i = 1; i <= Q; i++) {
    cout << ans[i] << "\n";
  }
}
