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;
template <typename T>
inline void ckmx(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void ckmi(T &x, T y) {
  x = min(x, y);
}
void read() {}
template <typename T, typename... Ts>
void read(T &x, Ts &...args) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  x *= f;
  read(args...);
}
int n, k, c[10010];
int ax[110], ay[110];
vector<int> ps[110];
int nw[110], mx;
int main() {
  read(n, k);
  mx = (n + k - 2) / (k - 1);
  for (int i = 1; i <= n * k; ++i) {
    read(c[i]);
    ps[c[i]].push_back(i);
  }
  int cnt = 0, nc = 1;
  for (int i = 1; i <= n * k; ++i) {
    int x = c[i];
    if (!ax[x]) {
      if (nw[x] == nc) {
        ax[x] = ps[x][nw[x] - 1];
        ay[x] = ps[x][nw[x]];
        cnt++;
        if (cnt == mx) cnt = 0, nc++;
      }
      nw[x]++;
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d %d\n", ax[i], ay[i]);
  return 0;
}
