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  maxi=5e5+10;
const int lm= 5e5+10;;
pair<int , pair<int , int>>  pr[maxi] ;
int tree[maxi] ;
int p[lm] ;
int ans[maxi] ;
void update(int idx, int val)
{
while(idx < maxi)
{
tree[idx] += val;
idx += (idx & -idx);
}
}

int read(int idx)
{
int sum = 0;
while(idx > 0)
{
sum += tree[idx];
idx -= (idx & -idx);
}
return sum;
}

int main()
{
   int n ,q;
   cin>>n;
   cin>>q;
   int a[n+1] ;
   for(int i=1;i<=n;i++)cin>>a[i];
  
   int x =1;

   for(int i=1 ; i<=q ; i++)
   {
       scanf("%d %d" , &pr[i].second.first , &pr[i].first) ;
       pr[i].second.second = i ;
   }
   sort(pr+1 , pr+q+1)  ;
   memset(tree , 0 , sizeof(tree)) ;
   memset(p , -1 , sizeof(p)) ;
   for(int i=1 ; i<=n ; i++)
   {
       if(p[a[i]]!= -1)
       update(p[a[i]] , -1)  ;
       p[a[i]] = i ;
       update(i , 1) ;


   while(x<=q && pr[x].first == i)
   {
       ans[pr[x].second.second] = read(pr[x].first) - read(pr[x].second.first-1);

       x++;

   }
   }
   for(int i=1 ;i<=q ; i++)
   printf("%d\n", ans[i]) ;
}