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>
inline bool read(T &x) {
  x = 0;
  register char c = getchar();
  register bool f = 0;
  while (!isdigit(c)) {
    if (c == EOF) return false;
    f ^= c == '-', c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return true;
}
template <class T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 ^ 48);
}
template <class T>
inline void print(T x, char c) {
  print(x), putchar(c);
}
template <class T, class... S>
inline bool read(T &x, S &...y) {
  return read(x) && read(y...);
}
int tr[300005], n, ans[300005], a[300005], m;
void add(int i, int x) {
  while (i <= n) tr[i] += x, i += i & (-i);
}
int sum(int i) {
  int t = 0;
  while (i) t += tr[i], i -= i & (-i);
  return t;
}
vector<pair<int, int> > p[300005];
int main() {
  int l, r, mid, pos;
  read(n, m);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    a[i] = i - a[i];
  }
  for (int i = 1; i <= m; i++) {
    read(l, r);
    l++;
    r = n - r;
    p[r].push_back(make_pair(l, i));
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      l = 1, r = i, mid, pos = 0;
      while (l <= r) {
        mid = l + r >> 1;
        if (sum(mid) >= a[i])
          l = mid + 1, pos = mid;
        else
          r = mid - 1;
      }
      add(1, 1);
      add(pos + 1, -1);
    }
    for (int j = 0; j < p[i].size(); j++)
      ans[p[i][j].second] = sum(p[i][j].first);
  }
  for (int i = 1; i <= m; i++) print(ans[i], '\n');
}
