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;

const int N = 500100;
int n, q, c[N], l, r, qs[N], pos[N];
vector<pair<int, int>> ls[N];

int bit[N];
void add(int i, int d) {
  for (; i <= n; i += i & -i) {
    bit[i] += d;
  }
}
int prefix(int i) {
  int s = 0;
  for (; i > 0; i -= i & -i) {
    s += bit[i];
  }
  return s;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> q;
  for (int i = 1; i <= n; ++i) {
    cin >> c[i];
  }
  for (int i = 1; i <= q; ++i) {
    cin >> l >> r;
    ls[r].emplace_back(l, i);
  }
  for (int i = 1, s = 0; i <= n; ++i) {
    int& p = pos[c[i]];
    if (p) add(p, -1);
    else ++s;
    p = i;
    add(i, 1);
    for (auto [l, j] : ls[i]) {
      qs[j] = s - prefix(l - 1);
    }
  }
  for (int i = 1; i <= q; ++i) {
    cout << qs[i] << "\n";
  }
  return 0;
}
