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 pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll; 
typedef long double ld;
typedef pair<ll, ll> pll;

const ll N = 5e5 + 30, Mod = 1e9 + 7;
const ll SQ = 330;

ll n, q, c[N], Fen[N], last[N], l[N], r[N], ans[N];
vector<ll> vec[N];
void Add(ll x, ll val) {
	x ++;
	for (++ x; x < N; x += x & -x) Fen[x] += val;
}
ll Get(ll x) {
	ll res = 0;
	x ++;
	for (++ x; x; x ^= x & -x) res += Fen[x];
	return res;
}
ll Get(ll l, ll r) {
	return Get(r) - Get(l - 1);
}
int main () {
	ios::sync_with_stdio(0), cin.tie(0);
 	cin >> n >> q;
 	for (int i = 1; i <= n; i ++) {
 		cin >> c[i];
	 }
	 for (int i = 1; i <= q; i ++) {
		cin >> l[i] >> r[i];
		vec[r[i]].pb(i);
	 }
	 for (int i = 1; i <= n; i ++) {
	 	if (last[c[i]]) Add(last[c[i]], -1);
	 	last[c[i]] = i;
	 	Add(i, 1);
	 	for (auto id : vec[i]) {
	 		ans[id] = Get(l[id], N - 5);
	 		
		}
	 }
	 for (int i = 1; i <= q; i ++) {
	 	cout << ans[i] << '\n';
	 }
 	
}