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<iostream>
#include<cmath>
#include<algorithm>

#define ll long long

using namespace std;

const int N = 1e6 + 10;

int tr[N];
int n,m;
int last[N];
int ans[N];
struct node{
	int l,r,pos;
}nod[N];

int color[N];

int lowbit(int x)
{
	return x & -x;
}

void add(int x, int a)
{
	for(int i=x; i<=n; i+=lowbit(i)) tr[i]+=a;
}

int query(int x)
{
	int rul = 0;
	for(int i=x; i; i-=lowbit(i)) rul += tr[i];
	return rul;
}

bool cmp(node a, node b)
{
	return a.r < b.r;
}

int main()
{
	cin>>n>>m;
	for(int i=1; i<=n; i++) scanf("%d",&color[i]);
	for(int i=1; i<=m; i++)
	{
		scanf("%d %d",&nod[i].l,&nod[i].r);
		nod[i].pos = i;
	}
	sort(nod+1, nod+m+1, cmp);
	int next = 1;
	for(int i=1; i<=m; i++)
	{
		for(int j=next; j<=nod[i].r; j++)
		{
			if(last[color[j]]) add(last[color[j]], -1);
			last[color[j]] = j;
			add(j, 1);
		}
		next = nod[i].r+1;
		ans[nod[i].pos] = query(nod[i].r) - query(nod[i].l-1);
	}
	for(int i=1; i<=m; i++)
	{
		cout<<ans[i]<<endl;
	}
	return 0;
}