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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int arr[100005], a[100005], b[100005], n, m, tot, num, cnt[100005];
struct Node {
  int l, r, id, ans;
  Node(int _l = 0, int _r = 0, int _id = 0, int _ans = 0) {
    l = _l;
    r = _r;
    id = _id;
    ans = _ans;
  }
} ans[100005];
bool cmp1(Node a, Node b) {
  if (a.l / num != b.l / num) return a.l / num < b.l / num;
  return (a.l / num) & 1 ? a.r < b.r : a.r > b.r;
}
bool cmp2(Node a, Node b) { return a.id < b.id; }
int query(int k) {
  int l = 1, r = tot, ret;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (b[mid] >= k)
      ret = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  return ret;
}
int nowl = 2, nowr = 1, now = 0;
void add(int x) {
  cnt[x]++;
  if (cnt[x] == b[x])
    now++;
  else if (cnt[x] == b[x] + 1)
    now--;
}
void del(int x) {
  cnt[x]--;
  if (cnt[x] == b[x])
    now++;
  else if (cnt[x] == b[x] - 1)
    now--;
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) a[i] = (arr[i] = read());
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    if (i == 1 || a[i] != a[i - 1]) b[++tot] = a[i];
  }
  for (int i = 1; i <= n; i++) arr[i] = query(arr[i]);
  for (int i = 1; i <= m; i++) {
    ans[i].l = read();
    ans[i].r = read();
    ans[i].id = i;
  }
  num = sqrt(n);
  sort(ans + 1, ans + m + 1, cmp1);
  for (int i = 1; i <= m; i++) {
    while (ans[i].l < nowl) add(arr[--nowl]);
    while (ans[i].r > nowr) add(arr[++nowr]);
    while (ans[i].l > nowl) del(arr[nowl++]);
    while (ans[i].r < nowr) del(arr[nowr--]);
    ans[i].ans = now;
  }
  sort(ans + 1, ans + m + 1, cmp2);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i].ans);
  return 0;
}
