#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
const long long maxn = 3e5 + 10;
long long a[maxn];
long long pre[maxn];
long long f[maxn];
long long x[maxn], y[maxn];
long long in(long long l, long long r, long long x) { return x >= l && x <= r; }
void clr(long long l, long long r) {
  for (long long i = l, iend = r; i <= iend; ++i) pre[i] = 0;
}
signed main() {
  n = read(), k = read();
  long long siz = n * k;
  for (long long i = 1, iend = siz; i <= iend; ++i) a[i] = read();
  long long u = k - 1;
  long long l = 1, r = u;
  while (l <= n) {
    if (r > n) r = n;
    for (long long i = 1, iend = siz; i <= iend; ++i) {
      long long v = a[i];
      if (!in(l, r, v)) continue;
      if (f[v]) continue;
      if (!pre[v]) {
        pre[v] = i;
        continue;
      }
      x[v] = pre[v];
      y[v] = i;
      f[v] = 1;
      clr(l, r);
    }
    l += u;
    r += u;
  }
  for (long long i = 1, iend = n; i <= iend; ++i) {
    cout << x[i] << " " << y[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;
long long n, m, k;
long long read() {
  long long s = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = s * 10 + ch - '0';
    ch = getchar();
  }
  return s * f;
}
const long long maxn = 3e5 + 10;
long long a[maxn];
long long pre[maxn];
long long f[maxn];
long long x[maxn], y[maxn];
long long in(long long l, long long r, long long x) { return x >= l && x <= r; }
void clr(long long l, long long r) {
  for (long long i = l, iend = r; i <= iend; ++i) pre[i] = 0;
}
signed main() {
  n = read(), k = read();
  long long siz = n * k;
  for (long long i = 1, iend = siz; i <= iend; ++i) a[i] = read();
  long long u = k - 1;
  long long l = 1, r = u;
  while (l <= n) {
    if (r > n) r = n;
    for (long long i = 1, iend = siz; i <= iend; ++i) {
      long long v = a[i];
      if (!in(l, r, v)) continue;
      if (f[v]) continue;
      if (!pre[v]) {
        pre[v] = i;
        continue;
      }
      x[v] = pre[v];
      y[v] = i;
      f[v] = 1;
      clr(l, r);
    }
    l += u;
    r += u;
  }
  for (long long i = 1, iend = n; i <= iend; ++i) {
    cout << x[i] << " " << y[i] << '\n';
  }
}

```