The Little Elephant loves playing with arrays. He has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai. 

Additionally the Little Elephant has m queries to the array, each query is characterised by a pair of integers lj and rj (1 ≤ lj ≤ rj ≤ n). For each query lj, rj the Little Elephant has to count, how many numbers x exist, such that number x occurs exactly x times among numbers alj, alj + 1, ..., arj.

Help the Little Elephant to count the answers to all queries.

Input

The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 105) — the size of array a and the number of queries to it. The next line contains n space-separated positive integers a1, a2, ..., an (1 ≤ ai ≤ 109). Next m lines contain descriptions of queries, one per line. The j-th of these lines contains the description of the j-th query as two space-separated integers lj and rj (1 ≤ lj ≤ rj ≤ n).

Output

In m lines print m integers — the answers to the queries. The j-th line should contain the answer to the j-th query.

Examples

Input

7 2
3 1 2 2 3 3 7
1 7
3 4


Output

3
1
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, “/ STACK : 268435456”)
template <class T>
inline int size(const T& c) {
  return c.size();
}
template <class T>
inline int lenght(const T& c) {
  return c.length();
}
int n, m;
int a[100011];
vector<int> b;
int f[100011][450];
int cnt[100011];
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  memset(cnt, 0, sizeof cnt);
  for (int i = (1), _b = (n); i <= _b; i++) {
    scanf("%d", &a[i]);
    if (a[i] <= n) cnt[a[i]]++;
  }
  b.clear();
  for (int i = (1), _b = (n); i <= _b; i++)
    if (cnt[i] >= i) (b).push_back((i));
  memset(f[0], 0, sizeof f[0]);
  for (int i = (1), _b = (n); i <= _b; i++)
    for (int j = (0), _b = ((b.size()) - 1); j <= _b; j++)
      f[i][j] = f[i - 1][j] + (a[i] == b[j]);
  for (int i = (0), _b = ((m)-1); i <= _b; i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    int ret = 0;
    for (int j = (0), _b = ((b.size()) - 1); j <= _b; j++)
      ret += (f[v][j] - f[u - 1][j]) == b[j];
    printf("%d ", ret);
    printf("\n");
  }
  return 0;
}
