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(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
const int N=5e5+10;
int cnt[N],ans[N],a[N];
int n,q,bl;
int res;
struct node
{
    int l,r,p;

}e[N];
bool cmp(const node &a,const node &b)
{
    return (a.l/bl)==(b.l/bl) ? a.r<b.r :a.l>b.l;
}
void remove(int x)
{
    if((--cnt[a[x]])==0) res--;
}
void add(int x)
{
    if((++cnt[a[x]])==1) res++;
}
int main()
{
    scanf("%d %d",&n,&q);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
     bl=sqrt(n);
    for(int i=1,l,r;i<=q;i++)
    {
        scanf("%d %d",&l,&r);
        e[i]={l,r,i};
    }
  //  cout<<"--------------"<<endl;

    sort(e+1,e+1+q,cmp);
   // cout<<"*****"<<endl;
    res=0;
    int curL=1,curR=0;
    for(int i=1;i<=q;i++)
    {
        //cout<<i<<endl;
        int L=e[i].l,R=e[i].r;
        while(curL<L) remove(curL++);
        while(curL>L) add(--curL);
        while(curR<R) add(++curR);
        while(curR>R) remove(curR--);
        ans[e[i].p]=res;
    }
    for(int i=1;i<=q;i++) printf("%d\n",ans[i]);

}
