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 = 1e6 + 100;
int bit[MAXN];
void update(int idx, int val)
{
	while(idx < MAXN)
	{
		bit[idx]+= val;
		idx += idx & (-idx);
	}
}

int query(int idx)
{
	int sum = 0;
	while(idx)
	{
		sum += bit[idx];
		idx -= idx & (-idx);
	}
	return sum;
}
pair<int,pair<int,int> >que[MAXN];
int v[MAXN];
int last[2 * MAXN];
int ans[MAXN];
int main()
{
	memset(last,-1,sizeof(last));
	int n;
	scanf("%d",&n);
	int q;
	scanf("%d",&q);
  
  for(int i = 1; i <= n; i++)
	{
		scanf("%d",&v[i]);
	}
	for(int i = 0; i < q; i++)
	{	
		scanf("%d",&que[i].second.first);
		scanf("%d",&que[i].first);
		que[i].second.second = i;
	}
	sort(que,que + q);
	int qt = 0;
	for(int r = 1; r <= n; r++)
	{
		update(r,1);
		if(last[v[r]] != -1)
			update(last[v[r]], -1);
		last[v[r]] = r;
	
		while(qt < q && que[qt].first == r)
		{
			ans[que[qt].second.second] = query(que[qt].first) - query(que[qt].second.first - 1);
			qt++;
		}
		if(qt == q)
			break;
	}
	for(int i = 0; i < q; i++)
	{
		printf("%d\n",ans[i]);
	}
	return 0;
}