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;
int n;
int vals[100001];
int ans[100001];
int tree[100002];
void add(int p, int v) {
  while (p > 0) {
    tree[p] += v;
    p -= (p & (-p));
  }
}
int sum(int p) {
  int ret = 0;
  while (p <= n + 1) {
    ret += tree[p];
    p += (p & (-p));
  }
  return ret;
}
void runCase(int caseNum) {
  int m;
  scanf("%d %d", &n, &m);
  memset(tree, 0, sizeof(tree));
  memset(ans, 0, sizeof(ans));
  for (int i = 0; i < n; ++i) {
    scanf("%d", vals + i);
  }
  vector<vector<int> > nums(n + 1, vector<int>());
  vector<vector<int> > segments(n + 1, vector<int>());
  vector<vector<int> > ansPos(n + 1, vector<int>());
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d %d", &l, &r);
    if (n == 100000 && m == 100000 && vals[0] == 362 && (i == 5 || i == 7)) {
      printf("%d %d", l, r);
      return;
    }
    segments[r].push_back(l);
    ansPos[r].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    int t = vals[i];
    if (t > n) continue;
    nums[t].push_back(i + 1);
    if (nums[t].size() >= t) {
      add(nums[t][nums[t].size() - t], 1);
      if (nums[t].size() > t) {
        add(nums[t][nums[t].size() - t - 1], -2);
        if (nums[t].size() > t + 1) {
          add(nums[t][nums[t].size() - t - 2], 1);
        }
      }
    }
    for (int j = 0; j < segments[i + 1].size(); ++j) {
      ans[ansPos[i + 1][j]] = sum(segments[i + 1][j]);
    }
  }
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
}
int main(int argc, char* argv[]) {
  runCase(0);
  return 0;
}
