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 oo = 2147483647;
int a1[100005], a2[100005], a3[100005], d[100005], b[100005];
int sum = 0, ans[100005], num[100005], n, m;
bool comp(int i, int j) {
  return a3[i] < a3[j] || (a3[i] == a3[j] && a2[i] < a2[j]);
}
void init() {
  int tmp, i;
  scanf("%d%d", &n, &m);
  tmp = max(1, (int)sqrt((double)n));
  for (i = 1; i <= n; i++) scanf("%d", d + i);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", a1 + i, a2 + i);
    a3[i] = (a2[i] - a1[i]) / tmp;
    b[i] = i;
  }
  sort(b + 1, b + 1 + m, comp);
}
void add(int v) {
  if (v > n) return;
  if (num[v] == v) sum--;
  num[v]++;
  if (num[v] == v) sum++;
}
void mirus(int v) {
  if (v > n) return;
  if (num[v] == v) sum--;
  num[v]--;
  if (num[v] == v) sum++;
}
void work() {
  int i, j, t1, t2, t3, t4;
  for (i = a1[b[1]]; i <= a2[b[1]]; i++) {
    add(d[i]);
  }
  ans[b[1]] = sum;
  for (i = 2; i <= m; i++) {
    t1 = a1[b[i - 1]];
    t2 = a2[b[i - 1]];
    t3 = a1[b[i]];
    t4 = a2[b[i]];
    if (t1 < t3)
      for (j = t1; j <= t3 - 1; j++) mirus(d[j]);
    if (t1 > t3)
      for (j = t3; j <= t1 - 1; j++) add(d[j]);
    if (t2 < t4)
      for (j = t2 + 1; j <= t4; j++) add(d[j]);
    if (t2 > t4)
      for (j = t4 + 1; j <= t2; j++) mirus(d[j]);
    ans[b[i]] = sum;
  }
  for (i = 1; i <= m; i++) printf("%d\n", ans[i]);
}
int main() {
  init();
  work();
}
