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;
int n, q, a[300005], l[300005], r[300005], p[300005], c[300005], las[300005],
    f[300005], ans[300005];
void add(int x) {
  if (!x) return;
  for (; x <= n; x += (x & -x)) c[x]++;
}
int quy(int x) {
  int res = 0;
  for (; x; x -= (x & -x)) res += c[x];
  return res;
}
int bs(int k) {
  if (k < 0) return 0;
  int l = 1, r = n, res = 0;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (quy(n) - quy(mid - 1) >= k)
      l = mid + 1, res = mid;
    else
      r = mid - 1;
  }
  return res;
}
bool cmp(const int a, const int b) { return r[a] > r[b]; }
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= q; i++) scanf("%d%d", &l[i], &r[i]), p[i] = i;
  sort(p + 1, p + q + 1, cmp);
  r[0] = n;
  for (int i = 1; i <= q; i++) {
    for (int j = n - r[p[i - 1]] + 1; j <= n - r[p[i]]; j++)
      add(f[j] = a[j] == j ? j : bs(j - a[j]));
    ans[p[i]] = quy(n) - quy(l[p[i]]);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
