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 n,q;
int c[500005],pre[500005],ans[500005];
vector<pair<int,int>> qs[500005];
int bit[2000005];

int sum(int pos){
    int res=0;
    while(pos>0){
        res+=bit[pos];
        pos-=pos&-pos;
    }
    return res;
}

void add(int pos,int val){
    while(pos<=n){
        bit[pos]+=val;
        pos+=pos&-pos;
    }
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n>>q;
    for(int i=1;i<=n;i++)cin>>c[i];
    for(int i=1;i<=q;i++){
        int l,r;
        cin>>l>>r;
        qs[r].emplace_back(l,i);
    }
    for(int i=1;i<=n;i++){
        add(i,1);
        if(pre[c[i]])add(pre[c[i]],-1);
        pre[c[i]]=i;
        for(pair<int,int> &p:qs[i]){
            ans[p.second]=sum(i)-sum(p.first-1);
        }
    }
    for(int i=1;i<=q;i++)cout<<ans[i]<<'\n';

    return 0;
}