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>
#define st first
#define nd second

using namespace std;
const int MN=5e5+10;
int n, q, D[MN], t[MN], last[MN], ans[MN];
vector<pair<pair<int, int>, int> > que;

void akt(int ile, int x){
    if(x>n) return;
    D[x]+=ile;
    akt(ile, x+(x&(-x)));
}
int sum(int x){
    if(x==0) return 0;
    return D[x]+sum(x-(x&(-x)));
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin>>n>>q;
    for(int i=1; i<=n; ++i) cin>>t[i];
    for(int i=0; i<q; ++i){
        int l, r;
        cin>>l>>r;
        que.push_back({{r, l}, i});
    }
    for(int i=1; i<=n; ++i) last[i]=-1;
    sort(que.begin(), que.end());
    int wsk=0;
    for(int i=1; i<=n; ++i){
        if(last[t[i]]!=-1) akt(-1, last[t[i]]);
        last[t[i]]=i;
        akt(1, i);
        while(que[wsk].st.st==i&&wsk<q){
            auto p=que[wsk];
            ans[p.nd]=sum(p.st.st)-sum(p.st.nd-1);
            wsk++;
        }
        if(wsk>=q)break;
    }
    for(int i=0; i<q; ++i) cout<<ans[i]<<'\n';
    return 0;
}
