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;
const int maxn = 3e5 + 5;
int a[maxn];
int bit[maxn];
int ans[maxn];
int query(int idx) {
  int ans = 0;
  for (int i = idx; i >= 1; i -= (i & -i)) {
    ans += bit[i];
  }
  return ans;
}
int n;
void update(int idx, int val) {
  if (idx <= 0) {
    return;
  }
  for (int i = idx; i <= n; i += (i & -i)) {
    bit[i] += val;
  }
}
int findLargestIdx(int target, int maksIdx) {
  int kiri = 1, kanan = maksIdx, mid, ans = -1;
  while (kiri <= kanan) {
    mid = (kiri + kanan) / 2;
    if ((query(maksIdx) - query(mid - 1)) >= target) {
      ans = mid;
      kiri = mid + 1;
    } else {
      kanan = mid - 1;
    }
  }
  return ans;
}
int main() {
  int q, i, j;
  scanf("%d %d", &n, &q);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = i - a[i];
  }
  vector<vector<int> > listKiri, listIdx;
  listKiri.assign(n + 1, vector<int>());
  listIdx.assign(n + 1, vector<int>());
  for (i = 1; i <= q; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    int start = 1 + x, end = n - y;
    listKiri[end].push_back(start);
    listIdx[end].push_back(i);
  }
  for (int r = 1; r <= n; r++) {
    if (a[r] >= 0) {
      int largestIdx = findLargestIdx(a[r], r);
      update(largestIdx, 1);
    }
    for (i = 0; i < listKiri[r].size(); i++) {
      int kiri = listKiri[r][i];
      int idxQuery = listIdx[r][i];
      ans[idxQuery] = query(r) - query(kiri - 1);
    }
  }
  for (i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
};
