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 + 10;

struct node{
	int l, r, id;
	bool operator < (const node &W) const
	{
		return r < W.r;
	}
}e[N];

int n, q;
int a[N];
int last[N];
int c[N];
int st[N];
int lowbit(int x) {return x &-x;}
int ans[N];

void add(int x, int y)
{
	for (; x <= n; x += lowbit(x))
	c[x] += y;
}

int ask(int x)
{
	int res = 0;
	for (;x; x -= lowbit(x))
	res += c[x];
	return res;
}



int main()
{
	scanf("%d", &n);	scanf("%d", &q);
	for (int i = 1; i <= n; i ++)
	{
		int x;
		scanf("%d", &x);
		a[i] = x;
	}
	for (int i = 1; i <= q; i ++)
	{
		int l, r;
		scanf("%d%d", &l, &r);
		e[i].id = i, e[i].l = l, e[i].r = r;
	}
	sort(e + 1, e + q + 1);
	int cnt = 1;
	for (int i = 1; i <= q; i ++)
	{
		for (int j = cnt; j <= e[i].r; j ++)
		{
			if(last[a[j]])
			{
				add(last[a[j]], -1);
			}
			add(j, 1);
			last[a[j]] = j;
		}
		cnt = e[i].r + 1;
		ans[e[i].id] = ask(e[i].r) - ask(e[i].l - 1);
	 }
	 for (int i = 1; i <= q; i ++)
	  printf("%d\n", ans[i]);
 }