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;
unordered_map<int, int> C;
const int b_size = 400;
const int LIM = 1e5 + 1;
struct query {
  int l, r, id;
};
int n, m, cnt;
int a[LIM];
vector<query> Q;
vector<int> res;
unordered_set<int> S;
bool cmp(const query &A, const query &B) {
  return make_pair(A.l / b_size, A.r) < make_pair(B.l / b_size, B.r);
}
void Enter() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  Q.resize(m);
  res.resize(m);
  for (int i = 0; i < m; ++i) {
    cin >> Q[i].l >> Q[i].r;
    Q[i].id = i;
  }
}
void INS(int id) {
  if (C[a[id]] == a[id]) S.erase(a[id]);
  ++C[a[id]];
  if (C[a[id]] == a[id]) S.insert(a[id]);
}
void RMV(int id) {
  if (C[a[id]] == a[id]) S.erase(a[id]);
  --C[a[id]];
  if (C[a[id]] == a[id]) S.insert(a[id]);
}
void Solve() {
  sort(Q.begin(), Q.end(), cmp);
  int l = 1, r = 0;
  for (int i = 0; i < Q.size(); ++i) {
    cnt = 0;
    while (l > Q[i].l) INS(--l);
    while (r < Q[i].r) INS(++r);
    while (l < Q[i].l) RMV(l++);
    while (r > Q[i].r) RMV(r--);
    res[Q[i].id] = S.size();
  }
  for (int x : res) cout << x << '\n';
}
int main() {
  Enter();
  Solve();
}
