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 N = 5e5 + 10;

int n, q;
int a[N], res[N], bit[N];
vector<int> pos[N];
vector<pair<int, int>> query[N];

void update(int x, int p) { for (; x <= n; x += x & -x) bit[x] += p; }
int get(int x) { int ret = 0; for (; x; x -= x & -x) ret += bit[x]; return ret; }

int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        if (pos[a[i]].empty()) update(i, 1);
        pos[a[i]].push_back(i);
    }
    for (int i = 1; i <= n; i++)
        reverse(pos[i].begin(), pos[i].end());
    for (int i = 1; i <= q; i++) {
        int l, r;
        scanf("%d%d", &l, &r);
        query[l].push_back(make_pair(r, i));
    }
    for (int i = 1; i <= n; i++) {
        for (pair<int, int> p: query[i])
            res[p.second] = get(p.first);
        update(pos[a[i]].back(), -1);
        pos[a[i]].pop_back();
        if (pos[a[i]].size()) update(pos[a[i]].back(), 1);
    }
    for (int i = 1; i <= q; i++)
        printf("%d\n", res[i]);
}
