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 M = 1e5 + 5, INF = 0x3f3f3f3f, mod = 1073741824;
const int block_size = 200;
struct query {
  int l, r, block, idx;
  query(int L = 0, int R = 0, int id = 0) {
    l = L;
    r = R;
    block = L / block_size;
    idx = id;
  }
  bool operator<(const query& a) const {
    return block != a.block ? block < a.block : r > a.r;
  }
};
vector<query> Q;
int sum[M], A[M];
vector<int> B;
int ans[M];
int cnt = 0;
void add(int x) {
  if (sum[A[x]] == B[A[x]]) cnt--;
  sum[A[x]]++;
  if (sum[A[x]] == B[A[x]]) cnt++;
}
void remove(int x) {
  if (sum[A[x]] == B[A[x]]) cnt--;
  sum[A[x]]--;
  if (sum[A[x]] == B[A[x]]) cnt++;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  memset(sum, 0, sizeof(sum));
  for (int j = 1; j <= n; j++) {
    scanf("%d", &A[j]);
    B.push_back(A[j]);
  }
  sort(B.begin(), B.end());
  B.erase(unique(B.begin(), B.end()), B.end());
  ;
  for (int j = 1; j <= n; j++)
    A[j] = lower_bound(B.begin(), B.end(), A[j]) - B.begin();
  for (int j = 1; j <= m; j++) {
    int l, r;
    scanf("%d%d", &l, &r);
    Q.push_back(query(l, r, j));
  }
  sort(Q.begin(), Q.end());
  int l = 1, r = 0;
  cnt = 0;
  for (auto now : Q) {
    while (r < now.r) add(++r);
    while (l < now.l) remove(l++);
    while (r > now.r) remove(r--);
    while (l > now.l) add(--l);
    ans[now.idx] = cnt;
  }
  for (int j = 1; j <= m; j++) printf("%d\n", ans[j]);
  return 0;
}
