#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long mas = 0;
vector<pair<long long, long long>> ans(1000000);
void solve(vector<vector<long long>> poses) {
  vector<long long> all;
  for (long long i = 0; i < poses.size(); i++) {
    for (long long j = 0; j < k; j++) {
      all.push_back(poses[i][j]);
    }
  }
  sort(all.begin(), all.end());
  map<long long, long long> conv;
  for (long long i = 0; i < all.size(); i++) {
    conv[all[i]] = i;
  }
  long long len = all.size();
  for (long long i = 0; i < poses.size(); i++) {
    for (long long j = 0; j < k; j++) {
      poses[i][j] = conv[poses[i][j]];
    }
  }
  vector<long long> start(poses.size());
  vector<bool> used(poses.size());
  for (long long _ = 0; _ < poses.size(); _++) {
    long long val = 1000000;
    long long ind = -1;
    for (long long i = 0; i < poses.size(); i++) {
      if (used[i]) {
        continue;
      }
      if (poses[i][start[i] + 1] < val) {
        val = poses[i][start[i] + 1];
        ind = i;
      }
    }
    ans[mas + ind].first = all[poses[ind][start[ind]]] + 1;
    ans[mas + ind].second = all[poses[ind][start[ind] + 1]] + 1;
    used[ind] = true;
    for (long long i = 0; i < poses.size(); i++) {
      if (used[i]) {
        continue;
      }
      if (poses[i][start[i]] < poses[ind][start[ind] + 1]) {
        start[i]++;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  vector<vector<long long>> poses(n);
  for (long long i = 0; i < n * k; i++) {
    long long c;
    cin >> c;
    poses[c - 1].push_back(i);
  }
  for (long long i = 0; i < n; i += k - 1) {
    vector<vector<long long>> kek;
    for (long long j = i; j < min(i + k - 1, n); j++) {
      kek.push_back(poses[j]);
    }
    mas = i;
    solve(kek);
  }
  for (long long i = 0; i < n; i++) {
    cout << ans[i].first << " " << ans[i].second << '\n';
  }
}

### Prompt

In CPP, your task is to solve 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, k;
long long mas = 0;
vector<pair<long long, long long>> ans(1000000);
void solve(vector<vector<long long>> poses) {
  vector<long long> all;
  for (long long i = 0; i < poses.size(); i++) {
    for (long long j = 0; j < k; j++) {
      all.push_back(poses[i][j]);
    }
  }
  sort(all.begin(), all.end());
  map<long long, long long> conv;
  for (long long i = 0; i < all.size(); i++) {
    conv[all[i]] = i;
  }
  long long len = all.size();
  for (long long i = 0; i < poses.size(); i++) {
    for (long long j = 0; j < k; j++) {
      poses[i][j] = conv[poses[i][j]];
    }
  }
  vector<long long> start(poses.size());
  vector<bool> used(poses.size());
  for (long long _ = 0; _ < poses.size(); _++) {
    long long val = 1000000;
    long long ind = -1;
    for (long long i = 0; i < poses.size(); i++) {
      if (used[i]) {
        continue;
      }
      if (poses[i][start[i] + 1] < val) {
        val = poses[i][start[i] + 1];
        ind = i;
      }
    }
    ans[mas + ind].first = all[poses[ind][start[ind]]] + 1;
    ans[mas + ind].second = all[poses[ind][start[ind] + 1]] + 1;
    used[ind] = true;
    for (long long i = 0; i < poses.size(); i++) {
      if (used[i]) {
        continue;
      }
      if (poses[i][start[i]] < poses[ind][start[ind] + 1]) {
        start[i]++;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  vector<vector<long long>> poses(n);
  for (long long i = 0; i < n * k; i++) {
    long long c;
    cin >> c;
    poses[c - 1].push_back(i);
  }
  for (long long i = 0; i < n; i += k - 1) {
    vector<vector<long long>> kek;
    for (long long j = i; j < min(i + k - 1, n); j++) {
      kek.push_back(poses[j]);
    }
    mas = i;
    solve(kek);
  }
  for (long long i = 0; i < n; i++) {
    cout << ans[i].first << " " << ans[i].second << '\n';
  }
}

```