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 N = 100005;
int n, m, second, a[N];
struct query {
  int l, r, p, block;
} t[N];
int res[N], curL, curR, curres = 0, cnt[N];
bool cmp(query a, query b) {
  return ((a.block < b.block) || (a.block == b.block && a.r < b.r));
}
void add(int pos) {
  if (a[pos] > n) return;
  cnt[a[pos]]++;
  if (cnt[a[pos]] == a[pos]) curres++;
  if (cnt[a[pos]] == a[pos] + 1) curres--;
}
void rem(int pos) {
  if (a[pos] > n) return;
  cnt[a[pos]]--;
  if (cnt[a[pos]] == a[pos]) curres++;
  if (cnt[a[pos]] == a[pos] - 1) curres--;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  second = sqrt(n);
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) {
    cin >> t[i].l >> t[i].r;
    t[i].p = i;
    t[i].block = t[i].l / second;
  }
  sort(t + 1, t + 1 + m, cmp);
  add(++curL);
  curR = 1;
  for (int i = 1; i <= m; i++) {
    while (curL < t[i].l) rem(curL++);
    while (curL > t[i].l) add(--curL);
    while (curR > t[i].r) rem(curR--);
    while (curR < t[i].r) add(++curR);
    res[t[i].p] = curres;
  }
  for (int i = 1; i <= m; i++) cout << res[i] << "\n";
  return 0;
}
