#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int mod = 1e9 + 7;
int t, n, k, a[MAX], s[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    int x = n, y = n;
    n *= k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<int> sol1(x), sol2(x);
    int cap = (x + k - 2) / (k - 1);
    for (int i = 1; i <= n; i++) {
      int top = s[i];
      int pos = 0;
      if (!y) break;
      if (sol1[a[i] - 1]) continue;
      for (int j = i - 1; j >= 1; j--) {
        top = max(top, s[j]);
        if (a[i] == a[j]) {
          if (top < cap) pos = j;
          break;
        }
      }
      if (pos) {
        y--;
        sol1[a[i] - 1] = pos;
        sol2[a[i] - 1] = i;
        for (int j = pos; j <= i; j++) s[j]++;
      }
    }
    for (int i = 0; i < x; i++) cout << sol1[i] << ' ' << sol2[i] << '\n';
  }
}

### Prompt

Your task is to create a CPP solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int mod = 1e9 + 7;
int t, n, k, a[MAX], s[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    int x = n, y = n;
    n *= k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    vector<int> sol1(x), sol2(x);
    int cap = (x + k - 2) / (k - 1);
    for (int i = 1; i <= n; i++) {
      int top = s[i];
      int pos = 0;
      if (!y) break;
      if (sol1[a[i] - 1]) continue;
      for (int j = i - 1; j >= 1; j--) {
        top = max(top, s[j]);
        if (a[i] == a[j]) {
          if (top < cap) pos = j;
          break;
        }
      }
      if (pos) {
        y--;
        sol1[a[i] - 1] = pos;
        sol2[a[i] - 1] = i;
        for (int j = pos; j <= i; j++) s[j]++;
      }
    }
    for (int i = 0; i < x; i++) cout << sol1[i] << ' ' << sol2[i] << '\n';
  }
}

```