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; 
typedef long long ll;
typedef long double ld;
#define pb push_back
#define fi first
#define se second
const ll mod = 1e9+7;
const int N = 5e5+5;
int bit[N];
void add(int idx){
	for(int i=idx;i>0;i-=(i&-i)){
		bit[i]++;
	}
}
int get(int idx){
	int ret = 0;
	for(int i=idx;i<N;i+=(i&-i)){
		ret+=bit[i];
	}
	return ret;
}
int a[N];
vector<int> qry[N];
int x[N];
int ans[N];
int prv[N];
void solve(){
	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;
		x[i] = l;
		ans[i]=  r-l+1;
		qry[r].pb(i);
	}
	for(int i=1;i<=n;i++){
		add(prv[a[i]]);
		for(int id:qry[i]){
			ans[id]-=get(x[id]);
		}
		prv[a[i]] = i;
	}
	for(int i=1;i<=q;i++){
		cout<<ans[i]<<endl;
	}
}
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int t=1; 
//	cin>>t; 
	while(t--){
		solve();
	}
	return 0;
}
