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 = 100010, MOD = 1000000007;
struct node {
  int l, r, id;
} a[N];
int sum[N], b[N], c[N], Ans[N], num[N];
int n, m, ans, sz;
bool cmp(node a, node b) {
  return a.l / sz != b.l / sz ? a.l / sz < b.l / sz : a.r < b.r;
}
void add(int i) {
  if (num[c[i]] == sum[i] - 1)
    ans++;
  else if (num[c[i]] == sum[i])
    ans--;
  num[c[i]]++;
}
void sub(int i) {
  if (num[c[i]] == sum[i])
    ans--;
  else if (num[c[i]] == sum[i] + 1)
    ans++;
  num[c[i]]--;
}
void solve() {
  sz = (int)sqrt(1.0 * n);
  memset(num, 0, sizeof num);
  sort(a + 1, a + 1 + m, cmp);
  int l = 1, r = 0;
  ans = 0;
  for (int i = 1; i <= m; i++) {
    while (r < a[i].r) add(++r);
    while (r > a[i].r) sub(r--);
    while (l < a[i].l) sub(l++);
    while (l > a[i].l) add(--l);
    Ans[a[i].id] = ans;
  }
  for (int i = 1; i <= m; i++) printf("%d\n", Ans[i]);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &sum[i]);
    c[i] = b[i] = sum[i];
  }
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; i++) {
    c[i] = lower_bound(b + 1, b + 1 + n, c[i]) - b;
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].l, &a[i].r);
    a[i].id = i;
  }
  solve();
  return 0;
}
