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 + 7;
int n;
struct query {
  int l, r;
  int id;
};
query qq[maxn];
bool cmp(const query& a, const query& b) { return a.r < b.r; }
int c[maxn], a[maxn], ans[maxn];
int lowbit(int x) { return x & (-x); }
void update(int i, int k) {
  while (i <= n) {
    c[i] += k;
    i += lowbit(i);
  }
}
int getsum(int i) {
  int res = 0;
  while (i > 0) {
    res += c[i];
    i -= lowbit(i);
  }
  return res;
}
int find_kth(int k) {
  int ans = 0, cnt = 0;
  for (int i = 20; i >= 0; i--) {
    ans += (1 << i);
    if (ans >= n || cnt + c[ans] >= k) {
      ans -= (1 << i);
    } else {
      cnt += c[ans];
    }
  }
  return ans + 1;
}
int main() {
  int q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i] = i - a[i];
  }
  for (int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    qq[i].l = x + 1;
    qq[i].r = n - y;
    qq[i].id = i;
  }
  sort(qq + 1, qq + 1 + q, cmp);
  int index = 1;
  for (int r = 1; r <= n; r++) {
    if (a[r] >= 0) {
      int left, right;
      left = 1;
      right = r;
      while (left <= right) {
        int mid = (left + right) >> 1;
        if (getsum(mid) >= a[r])
          left = mid + 1;
        else
          right = mid - 1;
      }
      update(1, 1);
      update(left, -1);
    }
    while (index <= q && qq[index].r <= r) {
      ans[qq[index].id] = getsum(qq[index].l);
      index++;
    }
  }
  for (int i = 1; i <= q; i++) {
    printf("%d\n", ans[i]);
  }
  printf("\n");
}
