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
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")


#include<bits/stdc++.h>
using namespace std;

const int MX=500005;

int bit[MX],last[MX],a[MX],ans[MX];
vector <int> qr[MX],ord[MX];

void upd(int i,int n,int x)
{
    while(i<=n){
        bit[i]+=x;
        i+=i & -i;
    }
}

int qry(int i)
{
    int s=0;
    while(i>0){
        s+=bit[i];
        i-=i & -i;
    }
    return s;
}

int main()
{
    int n,q,l,r;
    scanf("%d %d",&n,&q);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=0;i<q;i++){
        scanf("%d %d",&l,&r);
        qr[r].push_back(l);
        ord[r].push_back(i); //ord[i][j] has the index of qr[i][j]
    }
    for(int i=1;i<=n;i++){
        int p=last[a[i]];
        if(p) upd(p,n,-1);
        upd(i,n,1),last[a[i]]=i;
        int m=qr[i].size();
        for(int j=0;j<m;j++){
            int l=qr[i][j],r=i;
            ans[ord[i][j]]=qry(r)-qry(l-1);
            //printf("%d %d %d",l,r,ans[ord[i][j]]);
        }
    }
    for(int i=0;i<q;i++) printf("%d\n",ans[i]);
    return 0;
}

