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 F first
#define S second
using namespace std;
int n, q;
int a[600005], l[600005], fen[600005], ans[600005];
void update(int i,int k){
	for(int j=i;j<600000;j+=j&(-j)) fen[j]+=k;
}
int query(int i){
	int ans=0;
	for(int j=i;j>0;j-=j&(-j)) ans+=fen[j];
	return ans;
}
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>q;
	memset(fen, 0, sizeof fen);
	for(int i=1;i<=n;i++){
		cin>>a[i];
		l[i]=-1;
	}
	vector<pair<pair<int, int>, int>>queries;
	for(int i=0;i<q;i++){
		int l, r;
		cin>>l>>r;
		queries.push_back({{r, l}, i});
	}
	sort(queries.begin(), queries.end());
	int idx=0;
	for(int i=1;i<=n;i++){
		if(l[a[i]] != -1) update(l[a[i]], -1);
		l[a[i]]=i;
		update(i, 1);
		while(idx<q && queries[idx].F.F==i){
			ans[queries[idx].S]=query(queries[idx].F.F)-query(queries[idx].F.S-1);
			idx++;
		}
	}
	for(int i=0;i<q;i++){
		cout<<ans[i]<<"\n";
	}
}