The numbers 1,   2,   ...,   n ⋅ k are colored with n colors. These colors are indexed by 1,   2,   ...,   n. For each 1 ≤ i ≤ n, there are exactly k numbers colored with color i.

Let [a,   b] denote the interval of integers between a and b inclusive, that is, the set \\{a,   a + 1,   ...,   b\}. You must choose n intervals [a_1,   b_1],   [a_2,   b_2],   ..., [a_n,   b_n] such that: 

  * for each 1 ≤ i ≤ n, it holds 1 ≤ a_i < b_i ≤ n ⋅ k; 
  * for each 1 ≤ i ≤ n, the numbers a_i and b_i are colored with color i; 
  * each number 1 ≤ x ≤ n ⋅ k belongs to at most \left⌈ (n)/(k - 1) \right⌉ intervals. 



One can show that such a family of intervals always exists under the given constraints.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100, 2 ≤ k ≤ 100) — the number of colors and the number of occurrences of each color.

The second line contains n ⋅ k integers c_1,   c_2,   ...,   c_{nk} (1 ≤ c_j ≤ n), where c_j is the color of number j. It is guaranteed that, for each 1 ≤ i ≤ n, it holds c_j = i for exactly k distinct indices j.

Output

Output n lines. The i-th line should contain the two integers a_i and b_i.

If there are multiple valid choices of the intervals, output any.

Examples

Input


4 3
2 4 3 1 1 4 2 3 2 1 3 4


Output


4 5
1 7
8 11
6 12

Input


1 2
1 1


Output


1 2


Input


3 3
3 1 2 3 2 1 2 1 3


Output


6 8
3 7
1 4

Input


2 3
2 1 1 1 2 2


Output


2 3
5 6

Note

In the first sample, each number can be contained in at most \left⌈ (4)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:

<image>

In the second sample, the only interval to be chosen is forced to be [1,   2], and each number is indeed contained in at most \left⌈ (1)/(2 - 1) \right⌉ = 1 interval.

In the third sample, each number can be contained in at most \left⌈ (3)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:

<image>
#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
using namespace std;
void db() {cerr << endl;}
template <typename T, typename ...U> void db(T a, U ...b) {
    cerr << a << ' ', db(b...);
}
auto SEED = chrono::steady_clock::now().time_since_epoch().count();
const int N = 1e4 + 1, inf = 1e9, mod = 998244353;
int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    int n, k;
    cin >> n >> k;
    int cnt[n * k]{};
    vector<int> v[n];
    for (int i = 0; i < n * k; i++){
        int c;
        cin >> c;
        v[--c].push_back(i);
    }
    for (int c = 0; c < n; c++){
        int tar = 1, tmp = inf;
        for (int i = 1; i < k; i++){
            int ma = 0;
            for (int j = v[c][i - 1]; j < v[c][i]; j++)
                ma = max(ma, cnt[j]);
            if (ma < tmp){
                tmp = ma;
                tar = i;
            }
        }
        cout << v[c][tar - 1] + 1 << ' ' << v[c][tar] + 1 << '\n';
        for (int i = v[c][tar - 1]; i < v[c][tar]; i++)
            cnt[i]++;
    }
}