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;
template <class T>
struct FenwickTree1D {
  vector<T> bit;
  int n;
  FenwickTree1D(int a) {
    this->n = a + 1;
    bit.assign(a + 1, 0);
  }
  FenwickTree1D(vector<T> a) : FenwickTree1D(a.size()) {
    for (size_t i = 0; i < a.size(); i++) add(i, a[i]);
  }
  T sum(int idx) {
    T ret = 0;
    for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
    return ret;
  }
  T sum(int l, int r) { return sum(r) - sum(l - 1); }
  void add(int idx, T delta) {
    for (++idx; idx < n; idx += idx & -idx) bit[idx] += delta;
  }
  void range_add(int l, int r, int val) {
    add(l, val);
    add(r + 1, -val);
  }
};
int a[300009];
FenwickTree1D<int> T(300009);
vector<pair<int, int>> query[300009];
int ans[300009];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  int i = 0;
  while (i++ < q) {
    int x, y;
    cin >> x >> y;
    query[n - y].push_back({x + 1, i});
  }
  for (i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      int lo = 1, hi = i, mid, maxx = 0;
      while (lo <= hi) {
        mid = (lo + hi) / 2;
        int val = T.sum(mid);
        if (val >= a[i]) {
          maxx = mid;
          lo = mid + 1;
        } else
          hi = mid - 1;
      }
      if (maxx) T.range_add(1, maxx, 1);
    }
    for (auto u : query[i]) ans[u.second] = T.sum(u.first);
  }
  for (i = 1; i <= q; i++) cout << ans[i] << '\n';
}
