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 ll long long
using namespace std;

const ll MAX = 1e6+5;

ll a[MAX],tree[MAX];
ll last[MAX],cnt[MAX],n;
vector<pair<int,int>> query[MAX];
 
ll accum(ll id){
	ll ret = 0;
	while(id>0){
		ret += tree[id];
		id -= (id & -id);
	}
	return ret;
}
 
ll que(ll l,ll r){
	return accum(r)-accum(l-1);
}
 
void add(ll id,ll aum){
	while(id<=n){
		tree[id]+=aum;
		id += (id & -id);
	}
}
 
ll ans[MAX];
 
int main(){
	for(ll i=0;i<MAX;i++) last[i]=-1;
	ll l,r,q;
	cin>>n>>q;
 
	for(ll i=1; i<n+1; i++) 
		cin>>a[i];
	for(ll i=0; i<q; i++){
		cin>>l>>r;
		query[r].push_back({l,i});
	}
	for(ll i=1; i<n+1; i++) {
		if (last[a[i]]!=-1){
			--cnt[last[a[i]]];
			add(last[a[i]],-1);}
		last[a[i]]=i;
		cnt[last[a[i]]]++;
		add(last[a[i]],1);
 
		for(pair<int,int> qq: query[i]){
			ll sum = que(qq.first,i);
			ans[qq.second] = sum;
		}
	}
	for(ll i=0;i<q; i++)
		cout<<ans[i]<<endl;
}