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 f[maxn];
struct seg {
  int sum[maxn << 2];
  void modify(int p, int l, int r, int x) {
    sum[p]++;
    if (l == r) return;
    int mid = l + r >> 1;
    if (x <= mid)
      modify(p << 1, l, mid, x);
    else
      modify(p << 1 | 1, mid + 1, r, x);
  }
  int query(int p, int l, int r, int x, int y) {
    if (l >= x && r <= y) return sum[p];
    int mid = l + r >> 1;
    if (mid >= y)
      return query(p << 1, l, mid, x, y);
    else if (x > mid)
      return query(p << 1 | 1, mid + 1, r, x, y);
    else
      return query(p << 1, l, mid, x, y) + query(p << 1 | 1, mid + 1, r, x, y);
  }
  int find(int p, int l, int r, int k) {
    if (l == r) return l;
    int mid = l + r >> 1;
    if (sum[p << 1 | 1] >= k)
      return find(p << 1 | 1, mid + 1, r, k);
    else
      return find(p << 1, l, mid, k - sum[p << 1 | 1]);
  }
} trp, trs;
struct point {
  int x, y;
  inline bool operator<(const point a) const { return y > a.y; }
} pt[maxn];
struct line {
  int l, r, id;
  inline bool operator<(const line a) const { return l > a.l; }
} ln[maxn];
int qr[maxn];
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (register int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x > i)
      f[i] = -1;
    else if (x == i)
      f[i] = i - 1;
    else {
      if (trp.sum[1] < i - x)
        f[i] = -1;
      else
        f[i] = trp.find(1, 0, n - 1, i - x);
    }
    if (f[i] != -1) trp.modify(1, 0, n - 1, f[i]);
    pt[i] = {i, f[i]};
  }
  for (register int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ln[i] = {x + 1, n - y, i};
  }
  sort(pt + 1, pt + 1 + n);
  sort(ln + 1, ln + 1 + q);
  for (register int i = 1, j = 1; i <= q; i++) {
    while (j <= n && pt[j].y >= ln[i].l - 1) {
      trs.modify(1, 1, n, pt[j].x);
      ++j;
    }
    qr[ln[i].id] = trs.query(1, 1, n, ln[i].l, ln[i].r);
  }
  for (register int i = 1; i <= q; i++) printf("%d\n", qr[i]);
  return 0;
}
