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>
#pragma GCC optimize("Ofast", "unroll-loops")
using namespace std;
const int inf = numeric_limits<int>::max() / 2;
const long double eps = 1e-9;
const long double pi = acos(-1);
int SQRT = 320;
int n, m, cur = 0;
vector<int> a, l, r, ind, ans, cnt;
inline bool comp(int i, int j) {
  if (i / SQRT != j / SQRT) {
    return l[i] < l[j];
  }
  return r[i] < r[j];
}
inline void add(int i) {
  if (a[i] <= n) {
    if (cnt[a[i]] == a[i]) {
      --cur;
    }
    ++cnt[a[i]];
    if (cnt[a[i]] == a[i]) {
      ++cur;
    }
  }
}
inline void del(int i) {
  if (a[i] <= n) {
    if (cnt[a[i]] == a[i]) {
      --cur;
    }
    --cnt[a[i]];
    if (cnt[a[i]] == a[i]) {
      ++cur;
    }
  }
}
inline void solve() {
  ind.resize(m);
  cnt.assign(n + 1, 0);
  iota((ind).begin(), (ind).end(), 0);
  sort((ind).begin(), (ind).end(), comp);
  ans.resize(m, 0);
  int cl = 0, cr = -1;
  for (int i : ind) {
    while (cl > l[i]) {
      add(cl - 1);
      --cl;
    }
    while (cr < r[i]) {
      add(cr + 1);
      ++cr;
    }
    while (cl < l[i]) {
      del(cl);
      ++cl;
    }
    while (cr > r[i]) {
      del(cr);
      --cr;
    }
    ans[i] = cur;
  }
  for_each((ans).begin(), (ans).end(), [](int el) { cout << el << "\n"; });
}
inline void read() {
  cin >> n >> m;
  a.resize(n + 1);
  SQRT = sqrt(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  l.resize(m), r.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i];
    --l[i], --r[i];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read();
  solve();
  return 0;
}
