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 N = 1e6 + 100;

int a[N], bit[N], vis[N], ans[N];

struct P{
	int l, r, pos;
}p[N];

void add(int pos, int val) {
	for (int i = pos; i <= N; i += i & (-i)) {
		bit[i] += val;
	}
}

int sum(int pos) {
	int ans = 0;
	for (int i = pos; i != 0; i -= i & (-i)) {
		ans += bit[i];
	}
	return ans;
}

int main() {
	int n, m; cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= m; i++) {
		cin >> p[i].l >> p[i].r;
		p[i].pos = i;
	}
	sort(p + 1, p + 1 + m, [&] (P x, P y) {
		return x.r < y.r;
	});
	int r = 1;
	for (int i = 1; i <= m; i++) {
		for (int j = r; j <= p[i].r; j++) {
			if (vis[a[j]])
				add(vis[a[j]], -1);
			add(j, 1);
			vis[a[j]] = j;
		}
		r = p[i].r + 1;
		ans[p[i].pos] = sum(p[i].r) - sum(p[i].l - 1);
	}
	for (int i = 1; i <= m; i++) {
		cout << ans[i] << "\n";
	}
}