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 <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int seg[1000005];
int n;
void upd(int i){
    ++seg[i+=n];
    while(i>1){
        i/=2;
        seg[i]= seg[2*i]+seg[2*i+1];
    }
}
int find(int l,int r){
    l+=n; r+=n;
    int res =0;
    while(l<r){
        if(l%2) res+=seg[l++];
        if(r%2) res+=seg[--r];
        l/=2; r/=2;
    }
    return res;
}
map<int,int> last;
vector<pair<int,int>> queries[500001];
int ans[500001];
int main(){
    int q; scanf("%d%d",&n,&q);
    int x[n];
    for(int i=0;i<n;i++) scanf("%d",&x[i]);
    for(int i=0;i<n;i++)  last[x[i]] = -1;
    for(int i=0;i<q;i++){
        int a,b; scanf("%d%d",&a,&b);
        --a; --b;
        queries[b].push_back({a,i});
    }
    for(int i=0;i<n;i++){
        if(last[x[i]]!=-1)
            upd(last[x[i]]);
        last[x[i]] = i;
        for(auto q : queries[i]){
            ans[q.second] = (i-q.first+1)-find(q.first,i);
        }
    }
    for(int i=0;i<q;i++)
        printf("%d\n",ans[i]);
}
