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 long long LINF = ~(((long long)0x1) << 63) / 2;
const int INF = 0X3F3F3F3F;
const double eps = 1e-7;
int a[111111];
const int n = 110000;
int lowbit(int t) { return t & (-t); }
int Sum(int end) {
  int sum = 0;
  while (end > 0) {
    sum += a[end];
    end -= lowbit(end);
  }
  return sum;
}
void Plus(int pos, int num) {
  while (pos <= n) {
    a[pos] += num;
    pos += lowbit(pos);
  }
}
vector<int> b[110000];
int ans[110000];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    int c;
    cin >> c;
    if (c > n) continue;
    b[c].push_back(i);
  }
  vector<pair<int, int> > pa;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    pa.push_back(make_pair(l, r));
  }
  for (int i = 1; i <= n; i++) {
    int pre = 0;
    int sum = 0;
    for (int j = 0; j < b[i].size(); j++) {
      sum++;
      if (sum == i) {
        int pos1, pos2, d1, d2;
        if (pre == 0)
          pos1 = 1;
        else
          pos1 = b[i][pre - 1] + 1;
        if (j == b[i].size() - 1)
          pos2 = n;
        else
          pos2 = b[i][j + 1] - 1;
        d1 = b[i][pre];
        d2 = b[i][j];
        for (int k = 0; k < pa.size(); k++) {
          int q1 = pa[k].first, q2 = pa[k].second;
          if (q1 <= d1 && q2 >= d2 && q1 >= pos1 && q2 <= pos2) ans[k]++;
        }
        sum--;
        pre++;
      }
    }
  }
  for (int i = 0; i < m; i++) cout << ans[i] << endl;
  return 0;
}
