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;
const int NAX = 2e5 + 5, MOD = 1000000007;
struct Solution {
  Solution() {}
  void solveCase();
};
struct Bit {
  int size;
  vector<int> table;
  Bit(int size) {
    this->size = size;
    table.resize(size);
  }
  void update(int i, int delta) {
    while (i < size) {
      table[i] += delta;
      i = i | (1 + i);
    }
  }
  int sum(int i) {
    int ret = 0;
    while (i >= 0) {
      ret += table[i];
      i = (i & (i + 1)) - 1;
    }
    return ret;
  }
  int rangeSum(int i, int j) {
    if (i == 0) return sum(j);
    return sum(j) - sum(i - 1);
  }
  void print(int off = 0) {}
};
void Solution::solveCase() {
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    --a[i];
    a[i] = i - a[i];
  }
  vector<int> Queries[n + 1];
  vector<int> ans(q), leftBound(q);
  for (size_t i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    leftBound[i] = x;
    Queries[n - y - 1].push_back(i);
  }
  int global = 0;
  Bit b(n);
  ;
  for (size_t r = 0; r < n; r++) {
    int targ = a[r];
    if (targ >= 0 && targ <= global) {
      int low = 0, high = r, ans = 0;
      while (low <= high) {
        int mid = low + high;
        mid /= 2;
        int currVal = global + b.sum(mid);
        if (currVal >= targ) {
          low = mid + 1;
          ans = max(ans, mid);
        } else
          high = mid - 1;
      }
      b.print(global);
      ;
      ++global;
      b.update(ans + 1, -1);
    }
    for (auto &iQ : Queries[r]) ans[iQ] = global + b.sum(leftBound[iQ]);
  }
  b.print(global);
  for (auto &x : ans) {
    cout << x << '\n';
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  Solution mySolver;
  for (int i = 1; i <= t; ++i) {
    mySolver.solveCase();
  }
  return 0;
}
