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;
bool myfunction(int i, int j) { return (i < j); }
int n, m;
int a[100001];
map<int, int> mp;
vector<int> st;
vector<int> cc[100001];
map<pair<int, int>, int> mpp;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    if (a[i] > n) continue;
    mp[a[i]]++;
  }
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); ++it) {
    int num = (*it).first, cnt = (*it).second;
    if (num <= cnt) st.push_back(num);
  }
  mp.clear();
  for (int i = 1; i <= n; ++i) {
    mp[a[i]]++;
    for (int j = 0; j < st.size(); ++j) {
      cc[i].push_back(mp[st[j]]);
    }
  }
  mp.clear();
  for (int i = 1; i <= m; ++i) {
    int l, r, ret = 0;
    scanf("%d%d", &l, &r);
    for (int j = 0; j < st.size(); ++j) {
      int hi = cc[r][j], lo;
      if (l == 1)
        lo = 0;
      else
        lo = cc[l - 1][j];
      if (hi - lo == st[j]) ++ret;
    }
    printf("%d\n", ret);
  }
  return 0;
}
