1552_E. Colors and Intervals_71
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>
using namespace std;
const long long N = 105, K = 105;
const long long oo = 1e18 + 7, mod = 1e9 + 7;
long long n, k, a[N * K];
bool ok[N * K], ok2[N * K];
long long lst[N * K];
pair<long long, long long> ans[N * K];
long long sum[N * K];
void process() {
  cin >> n >> k;
  for (long long i = 1; i <= n * k; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) ok[i] = 1;
  long long m = n;
  while (m) {
    for (long long i = 1; i <= n; i++) lst[i] = 0;
    long long ini = 0;
    for (long long i = 1; i <= n * k; i++) {
      if (!ok[a[i]]) continue;
      if (lst[a[i]]) {
        ini++;
        ans[a[i]] = {lst[a[i]], i};
        ok[a[i]] = 0;
        for (long long j = 1; j <= n; j++) lst[j] = 0;
        if (ini == min(m, (k - 1))) break;
      } else
        lst[a[i]] = i;
    }
    assert(ini == min(m, (k - 1)));
    m -= min(m, k - 1);
  }
  for (long long i = 1; i <= n; i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  process();
}
