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;
struct Que {
  int a, b;
  int id;
  inline void read(int i) {
    cin >> a >> b;
    id = i;
    return;
  }
  bool operator<(const Que &oth) const {
    if (a == oth.a) return id < oth.id;
    return a > oth.a;
  }
};
int p[300010];
int br[300010];
int tree[1100010];
int odp[300010];
Que que[300010];
pair<int, int> xd[300010];
int pot;
inline void add(int x) {
  for (x += pot - 1; x > 0; x /= 2) tree[x]++;
  return;
}
inline int read(int l, int r) {
  int ans = 0;
  for (l += pot - 1, r += pot - 1; l <= r; l /= 2, r /= 2) {
    if (l % 2 == 1) {
      ans += tree[l];
      l++;
    }
    if (r % 2 == 0) {
      ans += tree[r];
      r--;
    }
  }
  return ans;
}
inline int fnd(int x) {
  int i = 1;
  for (; i < pot;) {
    if (tree[2 * i + 1] >= x)
      i = 2 * i + 1;
    else {
      x -= tree[2 * i + 1];
      i = 2 * i;
    }
  }
  return i - pot;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, q, a, i, j;
  cin >> n >> q;
  for (pot = 1; pot <= n; pot *= 2)
    ;
  for (i = 1; i <= n; i++) {
    cin >> a;
    if (a > i)
      p[i] = n + 1;
    else
      p[i] = i - a;
    if (p[i] == n + 1)
      br[i] = 0;
    else if (p[i] == 0)
      br[i] = i;
    else
      br[i] = fnd(p[i]);
    add(br[i] + 1);
    xd[i] = {-br[i], i};
  }
  for (i = 1; i <= q; i++) que[i].read(i);
  sort(que + 1, que + q + 1);
  sort(xd + 1, xd + n + 1);
  for (i = 1; i < 2 * pot; i++) tree[i] = 0;
  for (i = 1, j = 1; i <= q; i++) {
    while ((j <= n) && (-xd[j].first > que[i].a)) {
      add(xd[j].second);
      j++;
    }
    odp[que[i].id] = read(que[i].a + 1, n - que[i].b);
  }
  for (i = 1; i <= q; i++) cout << odp[i] << "\n";
  return 0;
}
