#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int maxn = 1005;
void go() {
  int n, k;
  cin >> n >> k;
  vector<int> v[n + 1];
  for (int i = 1; i < n * k + 1; i++) {
    int a;
    cin >> a;
    v[a].push_back(i);
  }
  int fin[101] = {}, l[101], r[101];
  int iter = 0;
  while (1) {
    priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
    for (int i = 1; i < n + 1; i++) {
      if (!fin[i]) {
        vector<int> a{v[i][1], v[i][0], i, 1};
        pq.push(a);
      }
    }
    if (!pq.size()) break;
    int ok = 0;
    while (pq.size()) {
      auto now = pq.top();
      pq.pop();
      if (now[1] > ok) {
        fin[now[2]] = 1;
        l[now[2]] = now[1];
        r[now[2]] = now[0];
        ok = r[now[2]];
      } else {
        if (now[3] < k - 1) {
          now[3]++;
          now[0] = v[now[2]][now[3]];
          now[1] = v[now[2]][now[3] - 1];
          pq.push(now);
        }
      }
    }
  }
  for (int i = 1; i < n + 1; i++) {
    cout << l[i] << ' ' << r[i] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int c = 0;
  int t;
  if (!c) {
    t = 1;
  } else {
    cin >> t;
  }
  while (t--) {
    go();
  }
}

### Prompt

Generate 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 mod = 998244353;
const int maxn = 1005;
void go() {
  int n, k;
  cin >> n >> k;
  vector<int> v[n + 1];
  for (int i = 1; i < n * k + 1; i++) {
    int a;
    cin >> a;
    v[a].push_back(i);
  }
  int fin[101] = {}, l[101], r[101];
  int iter = 0;
  while (1) {
    priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
    for (int i = 1; i < n + 1; i++) {
      if (!fin[i]) {
        vector<int> a{v[i][1], v[i][0], i, 1};
        pq.push(a);
      }
    }
    if (!pq.size()) break;
    int ok = 0;
    while (pq.size()) {
      auto now = pq.top();
      pq.pop();
      if (now[1] > ok) {
        fin[now[2]] = 1;
        l[now[2]] = now[1];
        r[now[2]] = now[0];
        ok = r[now[2]];
      } else {
        if (now[3] < k - 1) {
          now[3]++;
          now[0] = v[now[2]][now[3]];
          now[1] = v[now[2]][now[3] - 1];
          pq.push(now);
        }
      }
    }
  }
  for (int i = 1; i < n + 1; i++) {
    cout << l[i] << ' ' << r[i] << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int c = 0;
  int t;
  if (!c) {
    t = 1;
  } else {
    cin >> t;
  }
  while (t--) {
    go();
  }
}

```