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 ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int MX = 300005;
struct BIT {
  long long bit[MX];
  void add(int x, long long val) {
    for (int i = x; i < MX; i += i & -i) {
      bit[i] += val;
    }
  }
  long long get(int x) {
    long long res = 0;
    for (int i = x; i > 0; i -= i & -i) {
      res += bit[i];
    }
    return res;
  }
} bit;
vector<pi> qrys[MX];
int ans[MX];
void ad(int l, int r) {
  l++, r++;
  bit.add(l, 1);
  bit.add(r + 1, -1);
}
int at(int i) {
  i++;
  return bit.get(i);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<ll> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = (i + 1) - a[i];
  }
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    int a = x, b = (n - y - 1);
    qrys[b].push_back(make_pair(a, i));
  }
  for (int i = 0; i < n; i++) {
    int lo = 0;
    int hi = i;
    int ret = -1;
    while (lo <= hi) {
      int mid = (lo + hi) / 2;
      if (a[i] >= 0 && a[i] <= at(mid)) {
        lo = mid + 1;
        ret = mid;
      } else {
        hi = mid - 1;
      }
    }
    if (ret >= 0) ad(0, ret);
    for (pi to : qrys[i]) {
      ans[to.second] = at(to.first);
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << '\n';
  }
}
