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 INF = 1E9;
const int MOD = 1E9 + 7;
int n, m, l, r, t;
vector<int> arr, sol, ans;
map<int, vector<int>> occr;
map<int, vector<pair<int, int>>> qrs;
void readit() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; arr.push_back(t), ++i) scanf("%d", &t);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &l, &r);
    qrs[r - 1].push_back({l - 1, i});
  }
}
void solvit() {
  readit();
  ("DONE reading\n");
  sol.resize(n, 0);
  ans.resize(m, 0);
  ("%d %d\n", n, m);
  for (int i = 0; i < n; ++i) {
    ("%d ", i);
    t = arr[i];
    ("%d - ", t);
    occr[t].push_back(i);
    ("%d inserted - ", occr[t].size());
    if (occr[t].size() >= t) sol[occr[t][(occr[t].size() - t)]] += 1;
    ("1 - ");
    if (occr[t].size() > t) sol[occr[t][(occr[t].size() - t - 1)]] -= 2;
    ("2 - ");
    if (occr[t].size() > t + 1) sol[occr[t][(occr[t].size() - t - 2)]] += 1;
    ("3 - ");
    sort(qrs[i].rbegin(), qrs[i].rend());
    ("sorted\n");
    int j = i, tans = 0;
    for (auto q : qrs[i]) {
      for (; j >= q.first; j--) tans += sol[j];
      ans[q.second] = tans;
    }
  }
  ("DONE preprocessing\n");
  for (int i = 0; i < m; ++i) printf("%d\n", ans[i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  const clock_t begin_time = clock();
  solvit();
  ("\n=====\nUsed: %d ms",
   (int)(float(clock() - begin_time) / 1000 * CLOCKS_PER_SEC));
  return 0;
}
