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>
constexpr int N(3e5 + 5);
int n, q, a[N], ans[N];
std::vector<std::pair<int, int> > g[N];
struct Node {
  int add, min;
} t[N << 2];
inline void pushup(int o) {
  t[o].min = std::min(t[o << 1].min, t[o << 1 | 1].min);
}
inline void pushdown(int o) {
  if (t[o].add) {
    t[o << 1].add += t[o].add, t[o << 1 | 1].add += t[o].add;
    t[o << 1].min += t[o].add, t[o << 1 | 1].min += t[o].add;
    t[o].add = 0;
  }
}
void update(int o, int l, int r, int x, int z) {
  if (t[o].min >= z && r <= x) {
    t[o].add++, t[o].min++;
    return;
  }
  if (l == r) return;
  pushdown(o);
  int mid = l + r >> 1;
  update(o << 1, l, mid, x, z);
  if (x > mid && t[o << 1].min >= z) update(o << 1 | 1, mid + 1, r, x, z);
  pushup(o);
}
int ask(int o, int l, int r, int x) {
  if (l == r) return t[o].min;
  pushdown(o);
  int mid = l + r >> 1;
  return x <= mid ? ask(o << 1, l, mid, x) : ask(o << 1 | 1, mid + 1, r, x);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]), a[i] = i - a[i];
  }
  for (int i = 1; i <= q; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x++, y = n - y;
    g[y].push_back(std::pair<int, int>(x, i));
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] >= 0) update(1, 1, n, i, a[i]);
    for (auto s : g[i]) ans[s.second] = ask(1, 1, n, s.first);
  }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
