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>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ")";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int mod = 1e9 + 7;
inline int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline int mul(int a, int b) { return a * 1ll * b % mod; }
inline int power(int x, long long y) {
  int res = 1;
  while (y) {
    if (y & 1) res = mul(res, x);
    x = mul(x, x);
    y >>= 1;
  }
  return res;
}
const int B = 200;
const int N = 3e5 + 100;
int A[N];
int tab[N];
int ans[N];
vector<pair<int, int> > query[N];
void upt(int i, int val) {
  while (i < N) {
    tab[i] += val;
    i += i & -i;
  }
}
int get(int i) {
  int sm = 0;
  while (i) {
    sm += tab[i];
    i -= i & -i;
  }
  return sm;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(25);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    A[i] -= i;
    A[i] *= -1;
    if (A[i] < 0) A[i] = N;
  }
  int l, r;
  for (int i = 1; i <= q; ++i) {
    cin >> l >> r;
    query[n - r].push_back(make_pair(l + 1, i));
  }
  for (int i = 1; i <= n; ++i) {
    int lo = 1, hi = i, res = -1;
    while (lo <= hi) {
      int m = (lo + hi) >> 1;
      int curr = get(m);
      if (curr >= A[i]) {
        res = m;
        lo = m + 1;
      } else {
        hi = m - 1;
      }
    }
    if (res != -1) {
      upt(1, 1);
      upt(res + 1, -1);
    }
    for (auto it : query[i]) {
      ans[it.second] = get(it.first);
    }
  }
  for (int i = 1; i <= q; ++i) {
    cout << ans[i] << "\n";
  }
  return 0;
}
