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;
typedef long long ll;
#define N 500005
ll arr[N] , bit[N] ,ans[N] ,vis[N];

struct kj {
  ll l,r,id;
};

bool comp(kj & a,kj & b){
   return (a.r < b.r);
}

void update(ll id,ll val,ll n){
  while(id<=n){
    bit[id]+=val;
    id+=(id&-id);
  }
}

ll ask(ll id){
  ll s=0;
  while(id>0){
    s+=bit[id];
    id-=(id &-id);
  }
  return s;
}

int main() {
  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
  ll n,k ,i,j;
  cin>>n>>k;
  for(i=0;i<n;i++)cin>>arr[i];
  kj query[k];

  for(i=0;i<k;i++){
    cin>>query[i].l >> query[i].r;
    query[i].id =i;
  }
  sort(query,query+k,comp);
     
  memset(vis,-1,sizeof(vis));

  ll cnt=0;
  for(i=0;i<n;i++){
    if(vis[arr[i]]!=-1)
     update(vis[arr[i]]+1 ,-1,n);
    
    vis[arr[i]]=i;
    update(i+1,1,n);

    while(cnt<k && query[cnt].r-1==i){
      ans[query[cnt].id] = ask( query[cnt].r) -ask( query[cnt].l-1);
      cnt++;
    }

  }

  for(i=0;i<k;i++)cout<<ans[i]<<"\n";

}   