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 a[500002];
int ans[500002];
int last[500002];
vector<pair<int, int> > v[500002];
int ft[500002];
void update(int i, int v){
    for(;i<=500000;i+=(i&-i)) ft[i] += v;
}
int query(int i){
    int ans = 0;
    for(;i>0;i-=(i&-i)) ans+=ft[i];
    return ans;
}
int main(){
    int n, q;
    scanf("%d%d", &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);
        v[r].emplace_back(make_pair(l, i));
    }

    for(int i=1;i<=n;i++){
        if(last[a[i]] == 0) update(i, 1);
        else{
            update(last[a[i]], -1);
            update(i, 1);
        }
        for(int j=0;j<v[i].size();j++){
            int l = v[i][j].first, idx = v[i][j].second;
            ans[idx] = query(i) - query(l-1);
        }
        last[a[i]] = i;
    }


    for(int i=1;i<=q;i++) printf("%d\n", ans[i]);
}