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 max6 = 1e6 + 6;
const int oo = 1e9 + 9;
const long long inf = 1e18 + 18;
pair<pair<int, int>, int> query[max6];
int a[max6];
int ocur[max6];
int n, m;
int block;
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  if (a.first.first / block != b.first.first / block)
    return a.first.first / block < b.first.first / block;
  return a.first.second < b.first.second;
}
int main() {
  scanf("%d %d", &n, &m);
  block = sqrt(n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= m; ++i)
    scanf("%d %d", &query[i].first.first, &query[i].first.second),
        query[i].second = i;
  sort(query + 1, query + m + 1, cmp);
  std::vector<int> ans(m + 5, 0);
  int l = query[1].first.first;
  int r = query[1].first.second;
  int res = 0;
  for (int i = l; i <= r; ++i)
    if (a[i] <= n) {
      ocur[a[i]]++;
      if (ocur[a[i]] == a[i]) res++;
      if (ocur[a[i]] == a[i] + 1) res--;
    }
  ans[query[1].second] = res;
  for (int i = 2; i <= m; ++i) {
    int id = query[i].second;
    int newl = query[i].first.first;
    int newr = query[i].first.second;
    if (newl > l) {
      for (int j = l; j < newl; ++j)
        if (a[j] <= n) {
          ocur[a[j]]--;
          if (ocur[a[j]] == a[j]) res++;
          if (ocur[a[j]] == a[j] - 1) res--;
        }
    } else if (newl < l)
      for (int j = l - 1; j >= newl; --j)
        if (a[j] <= n) {
          ocur[a[j]]++;
          if (ocur[a[j]] == a[j]) res++;
          if (ocur[a[j]] == a[j] + 1) res--;
        }
    if (newr > r) {
      for (int j = r + 1; j <= newr; ++j)
        if (a[j] <= n) {
          ocur[a[j]]++;
          if (ocur[a[j]] == a[j]) res++;
          if (ocur[a[j]] == a[j] + 1) res--;
        }
    } else if (newr < r)
      for (int j = r; j > newr; --j)
        if (a[j] <= n) {
          ocur[a[j]]--;
          if (ocur[a[j]] == a[j]) res++;
          if (ocur[a[j]] == a[j] - 1) res--;
        }
    ans[id] = res;
    l = newl, r = newr;
  }
  for (int i = 1; i <= m; ++i) printf("%d\n", ans[i]);
}
