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;
const int maxn = 3e5 + 5;
struct Chairman_Tree {
  int cnt, lson[maxn * 40], rson[maxn * 40], sum[maxn * 40];
  void clear() { cnt = 0; }
  int Newnode(int rt) {
    int nrt = ++cnt;
    sum[nrt] = sum[rt] + 1;
    lson[nrt] = lson[rt];
    rson[nrt] = rson[rt];
    return nrt;
  }
  void build(int &rt, int l, int r) {
    rt = ++cnt;
    int mid = (l + r) >> 1;
    if (l == r) return;
    build(lson[rt], l, mid);
    build(rson[rt], mid + 1, r);
  }
  int update(int rt, int l, int r, int p) {
    int nrt = Newnode(rt);
    if (l == r) return nrt;
    int mid = (l + r) >> 1;
    if (p <= mid)
      lson[nrt] = update(lson[rt], l, mid, p);
    else
      rson[nrt] = update(rson[rt], mid + 1, r, p);
    return nrt;
  }
  int query_1(int rt, int l, int r, int p) {
    if (l == r) return l;
    int mid = (l + r) >> 1;
    if (sum[rson[rt]] >= p)
      return query_1(rson[rt], mid + 1, r, p);
    else
      return query_1(lson[rt], l, mid, p - sum[rson[rt]]);
  }
  int query_2(int rt, int l, int r, int ql, int qr) {
    if (ql <= l && qr >= r) return sum[rt];
    int mid = (l + r) >> 1, ret = 0;
    if (ql <= mid) ret += query_2(lson[rt], l, mid, ql, qr);
    if (qr > mid) ret += query_2(rson[rt], mid + 1, r, ql, qr);
    return ret;
  }
} solve;
int n, q, root[maxn];
int main() {
  scanf("%d%d", &n, &q);
  solve.clear();
  solve.build(root[0], 0, n - 1);
  for (int i = 1; i <= n; ++i) {
    int temp;
    scanf("%d", &temp);
    temp = i - temp;
    if (temp < 0 || solve.sum[root[i - 1]] < temp) {
      root[i] = root[i - 1];
      continue;
    }
    int lim;
    if (temp == 0)
      lim = i - 1;
    else
      lim = solve.query_1(root[i - 1], 0, n - 1, temp);
    root[i] = solve.update(root[i - 1], 0, n - 1, lim);
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    printf("%d\n", solve.query_2(root[n - y], 0, n - 1, x, n - 1));
  }
  return 0;
}
