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;
int main(){
  int n,q; cin >> n >> q;
  vector<int> line(n),from_list(n),vec,ans(q);
  for(auto&e:line)cin >> e;
  map<int,int> mp;
  for(int i = 0; i < n; ++i){
    if(mp.count(line[i]) == 0) from_list[i] = -1;
    else from_list[i] = mp[line[i]];
    mp[line[i]] = i;
  }
  vector<pair<int,int>> queries(q);
  for(auto&e:queries){cin >> e.first >> e.second;--e.first;--e.second;}
  static vector<pair<int,int>> r_base[500000];
  for(int i = 0; i < q; ++i)
    r_base[queries[i].second].emplace_back(queries[i].first,i);
  for(int i = 0; i < n; ++i){
    auto it = lower_bound(vec.begin(),vec.end(),from_list[i]);
    if(it == vec.end() || *it != from_list[i])vec.insert(it,from_list[i]);
    for(auto e:r_base[i])
      ans[e.second] = i-e.first+1-(vec.end()-lower_bound(vec.begin(),vec.end(),e.first));
  }
  for(auto e:ans)cout << e << endl;
}