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;
int q, A[100008], n;
vector<int> ind;
map<int, vector<int> > index1;
inline int LSOne(int n) { return n & (-n); }
class FenwickTree {
  vector<int> ft;

 public:
  FenwickTree(int n) { ft.assign(n + 1, 0); }
  int rsq(int b) {
    int sum = 0;
    for (; b > 0; b = b - LSOne(b)) {
      sum += ft[b];
    }
    return sum;
  }
  int rsq(int a, int b) { return rsq(b) - (a == 1 ? 0 : rsq(a - 1)); }
  void adjust(int idx, int a) {
    for (; idx < (int)ft.size(); idx += LSOne(idx)) ft[idx] += a;
  }
};
struct Query {
  int l, r;
  int ans;
} Q[100008];
bool compare(int i, int j) { return Q[i].r < Q[j].r; }
int main() {
  int pres = 0;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  for (int i = 0; i < q; i++) {
    cin >> Q[i].l >> Q[i].r;
    ind.push_back(i);
  }
  sort(ind.begin(), ind.end(), compare);
  FenwickTree tr(n);
  for (int end = 1; end <= n; end++) {
    index1[A[end]].push_back(end);
    int sz = index1[A[end]].size();
    if (sz >= A[end]) {
      tr.adjust(index1[A[end]][sz - A[end]], 1);
    }
    if (sz > A[end]) {
      tr.adjust(index1[A[end]][sz - A[end] - 1], -2);
    }
    if (sz > A[end] + 1) {
      tr.adjust(index1[A[end]][sz - A[end] - 2], 1);
    }
    while (Q[ind[pres]].r == end) {
      Q[ind[pres]].ans = tr.rsq(Q[ind[pres]].l, Q[ind[pres]].r);
      pres++;
      if (pres >= q) break;
    }
    if (pres >= q) break;
  }
  for (int i = 0; i < q; i++) {
    cout << Q[i].ans << endl;
  }
  return 0;
}
