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 + 3;
int n, q, a[MAXN], f[MAXN];
struct node {
  int l, r, num;
} tre[MAXN << 2];
inline void build(int i, int l, int r) {
  tre[i].l = l, tre[i].r = r, tre[i].num = 0;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(i << 1, l, mid);
  build(i << 1 | 1, mid + 1, r);
}
inline int query(int i, int s) {
  if (tre[i].l == tre[i].r) {
    return tre[i].l - 1;
  }
  if (s > tre[i << 1].num) {
    s -= tre[i << 1].num;
    return query(i << 1 | 1, s);
  } else
    return query(i << 1, s);
}
inline void modify(int i, int l, int r) {
  if (tre[i].l > r || tre[i].r < l) return;
  if (tre[i].l == l && tre[i].r == r) {
    tre[i].num++;
    return;
  }
  modify(i << 1, l, r);
  modify(i << 1 | 1, l, r);
  tre[i].num = tre[i << 1].num + tre[i << 1 | 1].num;
}
struct edge {
  int xs, xe, y, num;
  int t;
  friend bool operator<(edge a, edge b) {
    if (a.y != b.y) return a.y < b.y;
    return a.t < b.t;
  }
} s[MAXN << 1];
int ncnt, ans[MAXN], sum[MAXN];
int lowbit(int x) { return x & -x; }
void change(int x, int delta) {
  for (; x <= n; x += lowbit(x)) sum[x] += delta;
}
int query1(int l, int r) {
  int tot = 0;
  l--;
  for (; r; r -= lowbit(r)) tot += sum[r];
  for (; l; l -= lowbit(l)) tot -= sum[l];
  return tot;
}
int main() {
  scanf("%d%d", &n, &q);
  build(1, 0, n + 1);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == i) f[i] = i - 1;
    if (a[i] > i) f[i] = -1;
    if (a[i] < i) f[i] = query(1, tre[1].num - i + a[i] + 1);
    modify(1, f[i] + 1, f[i] + 1);
  }
  for (int i = 1; i <= n; i += 1) {
    s[++ncnt].t = 1, s[ncnt].xs = s[ncnt].xe = i, s[ncnt].y = f[i];
  }
  for (int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    y = n - y;
    x++;
    s[++ncnt].t = 2;
    s[ncnt].xs = x, s[ncnt].xe = y, s[ncnt].y = n + 3, s[ncnt].num = i;
    s[++ncnt].t = 0;
    s[ncnt].xs = x, s[ncnt].xe = y, s[ncnt].y = x - 1, s[ncnt].num = i;
  }
  sort(s + 1, s + ncnt + 1);
  for (int i = 1; i <= ncnt; i++) {
    if (s[i].t == 1) {
      if (s[i].y == -1) continue;
      change(s[i].xe, 1);
    } else if (s[i].t == 2) {
      int tot = query1(s[i].xs, s[i].xe);
      ans[s[i].num] = tot - ans[s[i].num];
    } else {
      int tot = query1(s[i].xs, s[i].xe);
      ans[s[i].num] = tot;
    }
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
