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;

const int maxn = 500005;
int a[maxn], last[maxn], f[maxn], ans[maxn];

struct Query {
	int l, r, id;
	bool operator<(const Query &Q) const {
		return r < Q.r;
	}
};
Query b[maxn];

void upd(int id, int val) {
	while(id < maxn) f[id] += val, id += id&-id;
}
int sum(int id) {
	int res = 0;
	while(id > 0) res += f[id], id -= id&-id;
	return res;
}

signed main() {
	
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	
	int n, q;
	cin >> n >> q;
	
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	
	for(int i = 0; i < q; i++) {
		cin >> b[i].l >> b[i].r;
		b[i].id = i;
	}
	
	sort(b, b + q);
		
	for(int i = 0; i < maxn; i++) {
		last[i] = -1;
	}
	
	int idx = 0;
	for(int i = 1; i <= n; i++) {
		
		if(last[a[i]] != -1)
			upd(last[a[i]], -1);
		
		last[a[i]] = i;
		upd(i, 1);
		
		while(idx < q && b[idx].r == i) {
			ans[b[idx].id] = sum(b[idx].r) - sum(b[idx].l - 1);
			idx++;
		}
	}
	
	for(int i = 0; i < q; i++) {
		cout << ans[i] << "\n";
	}
}
