We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8


Output

1
2
2
1
2
2
6
3
3
3
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> a(n);
  for (int &x : a) cin >> x;
  using T = tuple<int, int, int>;
  vector<T> b(q);
  for (int i = 0; i < q; ++i) {
    int l, r;
    cin >> l >> r;
    b[i] = {l - 1, r - 1, i};
  }
  sort(b.begin(), b.end(), [](const T &x, const T &y) {
    int xl, xr, xi;
    int yl, yr, yi;
    tie(xl, xr, xi) = x;
    tie(yl, yr, yi) = y;
    xl /= 708;
    yl /= 708;
    if (xl != yl) return xl < yl;
    else return xr < yr;
  });

  int l = 0, r = 0;

  vector<int> cnt(n + 1);
  int ta = 0;
  auto add = [&](int i) {
    if (++cnt[a[i]] == 1) {
      ++ta;
    }
  };
  auto del = [&](int i) {
    if (--cnt[a[i]] == 0) {
      --ta;
    }
  };

  add(0);

  vector<int> ans(q);
  for (auto &x : b) {
    int ql, qr, qi;
    tie(ql, qr, qi) = x;
    while (ql < l) add(--l);
    while (qr > r) add(++r);
    while (ql > l) del(l++);
    while (qr < r) del(r--);
    ans[qi] = ta;
  }

  for (int x : ans) {
    cout << x << endl;
  }
  return 0;
}

