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;
#define N 500001
int dat[N];
void update(int i,int x){
  for(;i<N;i+=i&-i) dat[i]+=x;
}
int sum(int i){
  int s=0;
  for(;i>0;i-=i&-i)s+=dat[i];
  return s;
}
int rangesum(int l, int r){
  return sum(r)-sum(l-1);
}
  
int main()
{
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n,q;
  cin>>n>>q;
  vector<int> ql[n+1],qid[n+1];
  int a[n+1];
  for(int i=1;i<=n;i++)
    cin>>a[i];
  for(int i=1;i<=q;i++){
    int l,r;
    cin>>l>>r;
    ql[r].push_back(l);
    qid[r].push_back(i);
  }
  int ans[q+1],temp[n+1];
  for(int i=1;i<=n;i++) temp[i]=0;
  for(int i=1;i<=n;i++){
    if(temp[a[i]]!=0) update(temp[a[i]],-1);
    temp[a[i]]=i;
    update(i,1);
    for(int j=0;j<ql[i].size();j++){
      ans[qid[i][j]]=rangesum(ql[i][j],i);
    }
  }
  for(int i=1;i<=q;i++) cout<<ans[i]<<endl;
  return 0;
}
