description: 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>
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  int a[100005];
  vector<int> c[105];
  int ans[105][2];
  int idx[105], vidx[105], vis[105];
  void go(int gn, int kk) {
    if (gn == 0 || kk < 2) {
      return;
    }
    for (int i = (1); i <= (gn); ++i) {
      vis[i] = 0;
    }
    while (1) {
      int mn = 1e9, minidx;
      for (int i = (1); i <= (gn); ++i) {
        if (c[idx[i]][vidx[i]] < mn) {
          mn = c[idx[i]][vidx[i]];
          minidx = i;
        }
      }
      if (vis[minidx] != 0) {
        ans[idx[minidx]][0] = vis[minidx];
        ans[idx[minidx]][1] = mn;
        idx[minidx] = idx[gn];
        vidx[minidx] = vidx[gn];
        gn--;
        go(gn, kk - 1);
        return;
      } else {
        vis[minidx] = mn;
        vidx[minidx]++;
      }
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = (1); i <= (n); ++i) {
      c[i].clear();
    }
    for (int i = (1); i <= (n * k); ++i) {
      cin >> a[i];
      c[a[i]].push_back(i);
    }
    int lay = n / (k - 1) + ((n % (k - 1)) == 0 ? 0 : 1);
    for (int i = (1); i <= (lay); ++i) {
      int l = 0;
      for (int j = (1); j <= (k - 1); ++j) {
        if ((i - 1) * (k - 1) + j > n) {
          break;
        }
        l++;
        idx[j] = (i - 1) * (k - 1) + j;
        vidx[j] = 0;
      }
      go(l, k);
    }
    for (int i = (1); i <= (n); ++i) {
      cout << ans[i][0] << " " << ans[i][1] << "\n";
    }
    cout << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    solve();
    return 0;
  }
  