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;
#define ll long long
#define ld long double
#define oo 0x3f3f3f3f

const int N = 500005;
int A[N];
int BIT[N];

void Add(int c, int x = 1){
    for(;c < N; c+=c&-c)
        BIT[c]+=x;
}

int Sum(int c, int s=0){
    for(; c; c-=c&-c)
        s+=BIT[c];
    return s;
}

vector<array<int,2>>Q[N];
int res[N];
int lastPos[N];

int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n,q;
    cin>>n>>q;
    for(int i=1; i<=n; i++)cin>>A[i];
    for(int i=1; i<=q; i++){
        int l,r;
        cin>>l>>r;
        Q[r].push_back({l,i});
    }

    for(int i=1; i<=n; i++){
        if(lastPos[A[i]])Add(lastPos[A[i]], -1);
        lastPos[A[i]] = i;
        Add(i, 1);

        for(auto q : Q[i])
            res[q[1]] = Sum(i) - Sum(q[0]-1);
    }

    for(int i=1; i<=q; i++)
        cout<<res[i]<<"\n";
}
