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(2)
#include<bits/stdc++.h>
using namespace std;
int n,q,x,y,l=1,r,t,num,a[666666],p[2333333],ans[666666],dv;
struct node{int l,r,bel;}ls[666666];
int cmp(node a,node b){
    if(a.l/dv!=b.l/dv)return a.l/dv<b.l/dv;
    return a.r/dv<b.r/dv;
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>q;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=q;i++)
        cin>>x>>y,ls[i]=(node){x,y,i};
    dv=(int)(sqrt(n));
    sort(ls+1,ls+q+1,cmp);
    for(int i=1;i<=q;i++){
        while(ls[i].l<l)num+=!p[a[--l]]++;
        while(ls[i].l>l)num-=!--p[a[l++]];
        while(ls[i].r>r)num+=!p[a[++r]]++;
        while(ls[i].r<r)num-=!--p[a[r--]];
        ans[ls[i].bel]=num;
    }
    for(int i=1;i<=q;i++)cout<<ans[i]<<endl;
}