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;
typedef long long ll;
const int sz = 5e5+5;

typedef pair <int,int> pii;
#define x first
#define y second

int a[sz], col[sz], bit[sz], ans[sz];
vector <pii> quers[sz];

int query(int x) {
   int res = 0;
   while(x < sz) {
      res += bit[x];
      x += x & (-x);
   }
   return res;
}

void upd(int x, bool f) {
   while(x) {
      if(f) bit[x]++;
      else bit[x]--;
      x -= x & (-x);
   }
}

int main() {
   int n, q;
   cin >> n >> q;
   for(int i=1; i<=n; i++) scanf("%d", &a[i]);
   for(int i=1; i<=q; i++) {
      int l, r;
      scanf("%d %d", &l, &r);
      quers[r].emplace_back(l, i);
   }
   for(int i=1; i<=n; i++) {
      int c = a[i];
      if(col[c]) upd(col[c], 0);
      col[c] = i;
      upd(col[c], 1);
      for(auto it : quers[i]) ans[it.y] = query(it.x);
   }
   for(int i=1; i<=q; i++) printf("%d\n", ans[i]);
}
