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 N=1e6+10;
int n,m,a[N],tr[N],ans[N],vis[N];
struct node
{
    int l,r,num;
}p[N];
bool cmp(node s1,node s2)
{return s1.r<s2.r;}
void add(int i,int k)
{
    while(i<=n)
    {
        tr[i]+=k;
        i+=(i&-i);
    }
}
int sum(int i)
{
    int s=0;
    while(i)
    {
        s+=tr[i];
        i-=(i&-i);
    }
    return s;
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    for(int i=1;i<=m;i++)
    {
        cin>>p[i].l>>p[i].r;
        p[i].num=i;
    }
    sort(p+1,p+m+1,cmp);
    int L=1;
    for(int i=1;i<=m;i++)
    {
        int num=p[i].num;
        int l=p[i].l;
        int r=p[i].r;
        for(int j=L;j<=r;j++)
        {
            int x=a[j];
            if(vis[x])add(vis[x],-1);
            add(j,1);
            vis[x]=j;
        }
        L=r+1;
        ans[num]=sum(r)-sum(l-1);
    }
    for(int i=1;i<=m;i++)
        printf("%d\n",ans[i]);
    return 0;
}
