#include <bits/stdc++.h>
using namespace std;
const int max_n = 103, inf = 1000111222;
int m[max_n * max_n];
int cnt[max_n * max_n];
int n, k;
int mx;
int lst[max_n];
bool used[max_n];
pair<int, int> res[max_n];
int get_max(int from, int to) {
  int ans = cnt[from];
  for (int i = from + 1; i <= to; ++i) {
    ans = cnt[i] > ans ? cnt[i] : ans;
  }
  return ans;
}
void add1(int from, int to) {
  for (int i = from; i <= to; ++i) {
    ++cnt[i];
  }
}
vector<pair<int, int>> solve() {
  vector<pair<int, int>> ans;
  memset(used, 0, sizeof(used));
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i <= n; ++i) {
    lst[i] = -1;
    res[i] = make_pair(-1, -1);
  }
  for (int i = 0; i < n * k; ++i) {
    int cur = m[i];
    if (used[cur]) {
      continue;
    }
    int pr = lst[cur];
    lst[cur] = i;
    if (pr == -1) {
      continue;
    }
    int tp = get_max(pr, i);
    if (tp == mx) {
      continue;
    }
    used[cur] = true;
    res[cur] = make_pair(pr, i);
    add1(pr, i);
  }
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      return ans;
    }
  }
  for (int i = 0; i < n; ++i) {
    ans.push_back(res[i]);
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n * k; ++i) {
    cin >> m[i];
    --m[i];
  }
  mx = (n + k - 2) / (k - 1);
  vector<pair<int, int>> ans = solve();
  if (ans.empty()) {
    reverse(m, m + n * k);
    ans = solve();
    if (ans.empty()) {
      exit(228);
    }
    for (int i = 0; i < n; ++i) {
      ans[i].first = (n * k - 1 - ans[i].first);
      ans[i].second = (n * k - 1 - ans[i].second);
      swap(ans[i].first, ans[i].second);
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 max_n = 103, inf = 1000111222;
int m[max_n * max_n];
int cnt[max_n * max_n];
int n, k;
int mx;
int lst[max_n];
bool used[max_n];
pair<int, int> res[max_n];
int get_max(int from, int to) {
  int ans = cnt[from];
  for (int i = from + 1; i <= to; ++i) {
    ans = cnt[i] > ans ? cnt[i] : ans;
  }
  return ans;
}
void add1(int from, int to) {
  for (int i = from; i <= to; ++i) {
    ++cnt[i];
  }
}
vector<pair<int, int>> solve() {
  vector<pair<int, int>> ans;
  memset(used, 0, sizeof(used));
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i <= n; ++i) {
    lst[i] = -1;
    res[i] = make_pair(-1, -1);
  }
  for (int i = 0; i < n * k; ++i) {
    int cur = m[i];
    if (used[cur]) {
      continue;
    }
    int pr = lst[cur];
    lst[cur] = i;
    if (pr == -1) {
      continue;
    }
    int tp = get_max(pr, i);
    if (tp == mx) {
      continue;
    }
    used[cur] = true;
    res[cur] = make_pair(pr, i);
    add1(pr, i);
  }
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      return ans;
    }
  }
  for (int i = 0; i < n; ++i) {
    ans.push_back(res[i]);
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n * k; ++i) {
    cin >> m[i];
    --m[i];
  }
  mx = (n + k - 2) / (k - 1);
  vector<pair<int, int>> ans = solve();
  if (ans.empty()) {
    reverse(m, m + n * k);
    ans = solve();
    if (ans.empty()) {
      exit(228);
    }
    for (int i = 0; i < n; ++i) {
      ans[i].first = (n * k - 1 - ans[i].first);
      ans[i].second = (n * k - 1 - ans[i].second);
      swap(ans[i].first, ans[i].second);
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << "\n";
  }
  return 0;
}

```