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;
const int maxn=1e6+60;
int n,q;
int a[maxn],c[maxn],vis[maxn],num[maxn];
struct data
{
    int l,r,pos;
    bool friend operator<(const data& x,const data& y)
    {
        return x.r<y.r;
    }
}e[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void updata(int i,int k)
{
    while(i<=n)
    {
        c[i]+=k;
        i+=lowbit(i);
    }

}
int getsum(int i)
{
    int res=0;
    while(i>0)
    {
        res+=c[i];
        i-=lowbit(i);
    }
    return res;
}
int main()
{
    cin>>n>>q;
    for(int i=1;i<=n;++i) cin>>a[i];
    for(int i=1;i<=q;++i)  cin>>e[i].l>>e[i].r,e[i].pos=i;
    sort(e+1,e+q+1);
    int next=1;
    for(int i=1;i<=q;++i)
    {
        for(int j=next;j<=e[i].r;j++)
        {
            if(vis[a[j]]) updata(vis[a[j]],-1);
            updata(j,1);
            vis[a[j]]=j;
        }
         next=e[i].r+1;
            num[e[i].pos]=getsum(e[i].r)-getsum(e[i].l-1);
    }
    for(int i=1;i<=q;++i) cout<<num[i]<<endl;
    return 0;
}