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>
struct interval {
  int l, r;
  int operator<(const interval &i) const { return r < i.r; }
} is[10005];
int n, k, a[10005], tot, app[105][105], cnt[105], ci, lim, sel[105];
int sell[105], selr[105];
int main() {
  scanf("%d%d", &n, &k);
  tot = n * k;
  lim = (n + k - 2) / (k - 1);
  for (int i = 1; i <= tot; ++i) {
    scanf("%d", a + i);
    app[a[i]][++cnt[a[i]]] = i;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j < k; ++j) is[++ci] = (interval){app[i][j], app[i][j + 1]};
  std::sort(is + 1, is + ci + 1);
  memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= ci; ++i)
    if (!sel[a[is[i].l]]) {
      int can = 1;
      for (int j = is[i].l; j <= is[i].r; ++j) can &= cnt[j] < lim;
      if (can) {
        for (int j = is[i].l; j <= is[i].r; ++j) ++cnt[j];
        sell[a[is[i].l]] = is[i].l;
        selr[a[is[i].r]] = is[i].r;
        sel[a[is[i].l]] = 1;
      }
    }
  for (int i = 1; i <= n; ++i) printf("%d %d\n", sell[i], selr[i]);
  return 0;
}
