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 ll = long long;
const ll mod = 998244353;
const ll N = 3e5 + 5;
const int base = 400;
using ull = unsigned long long;
using namespace std;
ll n, m, t, k, T, a[N], dp[N], lab[N], u, v, b[N], d[N], tong, cnt, q, fe[N],
    ans;
vector<ll> adj[N];
vector<pair<ll, ll> > adt[N];
void add(ll id, ll val) {
  for (; id <= n; id += id & -id) fe[id] += val;
}
ll get(ll id) {
  ll total = 0;
  for (; id; id -= id & -id) total += fe[id];
  return total;
}
ll bs(ll v, ll lim) {
  ll pos = 0, total = 0;
  for (int i = log2(lim); i >= 0; i--) {
    if (pos + (1 << i) <= lim && total + fe[pos + (1 << i)] >= v) {
      total += fe[pos + (1 << i)];
      pos += (1 << i);
    }
  }
  return pos;
}
void sol() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = i - a[i];
  }
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    u = u + 1;
    v = n - v;
    adt[v].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] >= 0) {
      ll rt = bs(a[i], i);
      add(1, 1);
      add(rt + 1, -1);
    }
    for (pair<ll, ll> x : adt[i]) {
      d[x.second] = get(x.first);
    }
  }
  for (int i = 1; i <= m; i++) cout << d[i] << '\n';
}
int main() {
  if (fopen("pieaters"
            ".inp",
            "r")) {
    freopen(
        "pieaters"
        ".inp",
        "r", stdin);
    freopen(
        "pieaters"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int ntest;
  ntest = 1;
  while (ntest-- > 0) sol();
}
