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;
int a[300005];
int mn[300005];
int id[300005];
int tag[300005];
int sz;
int ans[300005], tot;
int c[300005];
int pre[300005];
int ps[300005], p[300005];
void add(int x) {
  if (x == 0 || x > n) return;
  for (; x <= n; x += ((x) & (-x))) c[x]++;
}
int getsum(int x) {
  int res = 0;
  if (x > n) x = n;
  for (; x > 0; x -= ((x) & (-x))) res += c[x];
  return res;
}
struct query {
  int l, r, id;
  bool operator<(const query b) const { return r < b.r; }
} ask[300005];
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  memset(ps, 0x3f3f3f3f, sizeof(ps));
  sz = sqrt(n);
  int mx = n / sz;
  memset(mn, 0x3f, sizeof(mn));
  for (int i = 1; i <= n; i++) {
    id[i] = i / sz;
    a[i] = i - a[i];
    if (a[i] < 0) a[i] = 0x3f3f3f3f;
    mn[id[i]] = min(mn[id[i]], a[i]);
  }
  while (1) {
    bool flag = 0;
    int pos = -1;
    for (int i = mx; i >= 0; i--) {
      if (flag) break;
      if (mn[i] - tag[i] <= 0) {
        mn[i] = 0x3f3f3f3f;
        for (int j = max(i * sz, 1); j < min((i + 1) * sz, n + 1); j++) {
          a[j] -= tag[i];
          if (a[j] < 0) a[j] = 0x3f3f3f3f;
          if (a[j] == 0) pos = j;
        }
        if (pos != -1) {
          flag = 1;
          ans[tot++] = pos;
          ps[pos] = tot;
          a[pos] = 0x3f3f3f3f;
          for (int j = i + 1; j <= mx; j++) tag[j]++;
          for (int j = pos + 1; j < min((i + 1) * sz, n + 1); j++) {
            a[j]--;
            if (a[j] < 0) a[j] = 0x3f3f3f3f;
          }
        }
        for (int j = i * sz; j < min((i + 1) * sz, n + 1); j++)
          mn[i] = min(mn[i], a[j]);
        tag[i] = 0;
      }
    }
    if (!flag) break;
  }
  memset(p, 0x3f, sizeof(p));
  for (int i = 1; i <= n; i++) {
    pre[i] = max(pre[i - 1], tot - ps[i] + 1);
    p[i] = min(p[i - 1], ps[i]);
    if (p[i] == 0x3f3f3f3f) p[i] = n + 1;
  }
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &ask[i].l, &ask[i].r);
    ask[i].id = i;
  }
  sort(ask, ask + q);
  int now = n;
  for (int i = 0; i < q; i++) {
    while (now >= n - ask[i].r + 1) {
      add(ps[now--]);
    }
    ans[ask[i].id] = tot - pre[ask[i].l] - getsum(p[ask[i].l] - 1);
  }
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
}
