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;
const int maxn = 100005;
int n, m, x[maxn];
set<int> ok;
vector<int> okiv;
int g[maxn][500];
map<int, int> o, rr;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &x[i]);
    o[x[i]] += 1;
  }
  for (map<int, int>::iterator it = o.begin(); it != o.end(); ++it) {
    if (it->first <= it->second) {
      ok.insert(it->first);
    }
  }
  for (set<int>::iterator it = ok.begin(); it != ok.end(); ++it) {
    okiv.push_back(*it);
    rr[*it] = (int)okiv.size() - 1;
  }
  for (int i = 1; i <= n; ++i) {
    copy(g[i - 1], g[i], g[i]);
    if (ok.count(x[i])) {
      g[i][rr[x[i]]] += 1;
    }
  }
  int l, r, ivs = (int)okiv.size();
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &l, &r);
    int ans = 0;
    for (int j = 0; j < ivs; ++j) {
      if (g[r][j] - g[l - 1][j] == okiv[j]) {
        ans += 1;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
