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;
const int inf = 1e9 + 7;
const int N = 1e6 + 1;
int t[N];
int c[N];
int last[N];
int get(int x) {
    int ans = 0;
    for (int i = x; i >= 0; i = (i&(i + 1)) - 1) ans += t[i];
    return ans;
}
void update(int pos, int add) {
    if (pos < 0) return;
    for (int i = pos; i < N; i = (i|(i + 1))) t[i] += add;
}
vector <pair <int, int> > q[N];
int ans[N];
signed main () {
    ios_base::sync_with_stdio(NULL);
    cin.tie(0); cout.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++) cin >> c[i];
    for (int i = 0; i < m; i++) {
        int l, r;
        cin >> l >> r; l--; r--;
        q[r].push_back({l, i});
    }
    memset(last, -1, sizeof last);
    for (int i = 0; i < n; i++) {
        update(last[c[i]], -1);
        last[c[i]] = i;
        update(last[c[i]], 1);
        for (auto it : q[i]) ans[it.second] = get(i) - get(it.first - 1);
    }
    for (int i = 0; i < m; i++) {
        cout << ans[i] << '\n';
    }
}
