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 N = 3e5 + 5;
int c[N], n;
int lowbit(int x) { return x & (-x); }
void add(int x, int val) {
  while (x <= n) {
    c[x] += val;
    x += lowbit(x);
  }
}
int sum(int x) {
  int ans = 0;
  while (x) {
    ans += c[x];
    x -= lowbit(x);
  }
  return ans;
}
vector<pair<int, int> > q[N];
int a[N], ans[N];
int main() {
  int x;
  cin >> n >> x;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  for (int i = 1; i <= x; i++) {
    int l, r;
    cin >> l >> r;
    l = l + 1;
    r = n - r;
    q[r].push_back(make_pair(l, i));
  }
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      int l = 1, r = i;
      int pos = 0;
      while (l <= r) {
        int mid = (l + r) / 2;
        if (sum(mid) >= a[i]) {
          l = mid + 1;
          pos = mid;
        } else
          r = mid - 1;
      }
      if (pos) {
        add(1, 1);
        add(pos + 1, -1);
      }
    }
    for (int j = 0; j < q[i].size(); j++) {
      int l = q[i][j].first;
      int id = q[i][j].second;
      ans[id] = sum(l);
    }
  }
  for (int i = 1; i <= x; i++) {
    cout << ans[i] << endl;
  }
}
