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;
using ull = unsigned long long;
using ll = long long;
int n, q;
int a[300005];
int ans[300005];
vector<pair<int, int> > e[300005];
int f[300005];
void add(int i, int x) {
  for (; i <= n; i += i & -i) {
    f[i] += x;
  }
}
int get(int i) {
  int v = 0;
  for (; i > 0; i -= i & -i) {
    v += f[i];
  }
  return v;
}
int get_idx(int x, int bd) {
  int l = 0, r = bd + 1, m;
  while (r - l > 1) {
    m = (l + r) >> 1;
    if (get(m) >= x) {
      l = m;
    } else {
      r = m;
    }
  }
  return l;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  for (int i = 1, x, y; i <= q; ++i) {
    cin >> x >> y;
    e[n - y].push_back({i, x + 1});
  }
  int cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (a[i] >= 0) {
      int idx = get_idx(a[i], i);
      add(1, 1), add(idx + 1, -1);
      ++cnt;
    }
    for (auto x : e[i]) {
      ans[x.first] = get(x.second);
    }
  }
  for (int i = 1; i <= q; ++i) {
    cout << ans[i] << endl;
  }
  return 0;
}
