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 pair<int,int> pii;
const int MAXN=500005;
vector<pii> que[MAXN];
int n,a[MAXN],last[MAXN],bit[MAXN],ans[MAXN];

void update(int now,int val){
	for(int i=now;i<=n;i+=(i&(-i)))bit[i]+=val;
}

int query(int now){
	int ret=0;
	for(int i=now;i>0;i-=(i&(-i)))ret+=bit[i];
	return ret;
}

int main(){
	int q,l,r;
	cin >> n >> q;
	for(int i=1;i<=n;i++)cin >> a[i];
	for(int i=1;i<=q;i++){
		cin >> l >> r;
		que[r].push_back({l,i});
	}
	for(int i=1;i<=n;i++){
		if(last[a[i]]!=0)update(last[a[i]],-1);
		last[a[i]]=i;
		update(i,1);
		for(auto isi : que[i]){
			ans[isi.second]=query(i)-query(isi.first-1);
		}
	}
	for(int i=1;i<=q;i++)cout << ans[i] << '\n';
	return 0;
}