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>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb emplace_back
#define sz(x) (ll)x.size()
typedef long long ll;
using namespace std;

const int N = 500001;

int ft[N];

void upd(int i, int x){
	for(; i < N; i += i&(-i)){
		ft[i] += x;
	}
}

int get(int i){
	int val = 0;
	for(; i > 0; i -= i&(-i)){
		val += ft[i];
	}
	return val;
}

int n, q, c[N];
int last[N];
vector<int> vl[N], vi[N]; 
int ans[N];

int main(){
    IOS

    cin >> n >> q;
    for(int  i = 1; i <= n; i++){
    	cin >> c[i];
    }
    for(int i = 1; i <= q; i++){
    	int l, r;
    	cin >> l >> r;
    	vl[r].pb(l);
    	vi[r].pb(i);
    }
    for(int i = 1; i <= n; i++){
    	if(last[c[i]] != 0)
    		upd(last[c[i]], -1);
    	upd(i, 1);
    	last[c[i]] = i;
    	for(int j = 0; j < sz(vl[i]); j++){
    		ans[vi[i][j]] = get(i) - get(vl[i][j] - 1);
    	}
    }
    for(int i = 1; i <= q; i++){
    	cout << ans[i] << endl;
    }

    return 0;
}