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 <cstdio>
#include <vector>
#include <utility>

using namespace std;

typedef pair<int,int> pii;
const int N=500005;

int n,m;
int c[N],pos[N],ans[N];
vector<pii> q[N];
int tr[N];

void add(int x,int c)
{
    for(;x<=n;x+=x&-x)tr[x]+=c;
}

int sum(int x)
{
    int res=0;
    for(;x;x&=x-1)res+=tr[x];
    return res;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",c+i);
    for(int i=0;i<m;i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        q[r].push_back(make_pair(l,i));
    }
    for(int i=1;i<=n;i++)
    {
        if(pos[c[i]]>0)add(pos[c[i]],-1);
        add(i,1);
        pos[c[i]]=i;
        for(int j=0;j<q[i].size();j++)
        {
            pii it=q[i][j];
            int l=q[i][j].first,id=q[i][j].second;
            ans[id]=sum(i)-sum(l-1);
        }
    }
    for(int i=0;i<m;i++)
        printf("%d\n",ans[i]);
    return 0;
}