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;
class Fenwick {
  vector<int> x;
  int total;

 public:
  Fenwick(int n) {
    x = vector<int>(n + 1);
    total = 0;
  }
  int sum_to(size_t n) {
    int ans = 0;
    int i = 1;
    while (n > 0) {
      if (n & i) {
        ans += x[n];
        n -= (n & i);
      }
      i <<= 1;
    }
    return ans;
  }
  int sum_from(size_t n) { return total - sum_to(n); }
  void increment(int n) {
    int i = 1;
    while (n < x.size()) {
      if (n & i) {
        x[n]++;
        n += (n & i);
      }
      i <<= 1;
    }
    total += 1;
  }
  void print() {
    for (size_t i = 0; i < x.size(); i++) {
      cout << sum_from(i) << " ";
    }
    cout << endl;
  }
};
int main() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = i + 1 - a[i];
  }
  map<int, vector<pair<int, int>>> query;
  vector<int> ans(q);
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    query[n - y].push_back(pair<int, int>(i, x));
  }
  Fenwick weight(n);
  for (int i = 0; i < n; i++) {
    if (a[i] >= 0 && weight.sum_from(0) >= a[i]) {
      int upper = i + 1;
      int lower = 0;
      while (lower + 1 < upper) {
        int mid = (lower + upper) / 2;
        if (weight.sum_from(mid) >= a[i]) {
          lower = mid;
        } else {
          upper = mid;
        }
      }
      weight.increment(lower + 1);
    }
    for (pair<int, int> q : query[i + 1]) {
      ans[q.first] = weight.sum_from(q.second);
    }
  }
  for (int i : ans) {
    cout << i << "\n";
  }
}
