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 x first
#define y second
#define all(v) v.begin(), v.end()
#define compress(v) sort(all(v)), v.erase(unique(all(v)), v.end())
using namespace std;

typedef long long ll;
typedef pair<int, int> p;

struct Query{
    int s, e, x;
    bool operator < (const Query &t) const {
        return p(s/700, e) < p(t.s/700, t.e);
    }
} qry[505050];

int n, q, a[505050];
int cnt[505050], now, res[505050];

void insert(int x){ if(!cnt[a[x]]++) now++; }
void erase(int x){ if(!--cnt[a[x]]) now--; }

int main(){
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    cin >> n >> q;
    for(int i=1; i<=n; i++) cin >> a[i];
    for(int i=1; i<=q; i++) cin >> qry[i].s >> qry[i].e, qry[i].x = i;
    sort(qry+1, qry+q+1);
    int s = 1, e = 1; insert(1);
    for(int i=1; i<=q; i++){
        while(qry[i].s < s) insert(--s);
        while(e < qry[i].e) insert(++e);
        while(s < qry[i].s) erase(s++);
        while(qry[i].e < e) erase(e--);
        res[qry[i].x] = now;
    }
    for(int i=1; i<=q; i++) cout << res[i] << "\n";
}