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;
  const long long INF_LL = 4e18;
  const int INF = 1e9 + 100;
  const int MOD = 998244353;
  const long double EPS = 1e-10;
  const int BLOCK = 550;
  const int dx[4] = {-1, 0, 1, 0};
  const int dy[4] = {0, 1, 0, -1};
  mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
  void open_file() {}
  const int maxN = 2e6 + 100;
  void sol() {
    int n, k;
    cin >> n >> k;
    vector<int> c(n * k);
    for (int &x : c) {
      cin >> x;
      x -= 1;
    }
    vector<int> used(n * k, 0);
    vector<int> last(n * k, -1);
    vector<int> dp(n * k, 0);
    vector<pair<int, int> > ans(n);
    for (int i = 0; i < n * k; i++) {
      int x = c[i];
      if (last[x] == -1) {
        last[x] = i;
        if (i == 0) continue;
        dp[i] = dp[i - 1];
      } else {
        if (used[x] == 0) {
          if (dp[i - 1] - dp[last[x]] < (n + k - 2) / (k - 1)) {
            dp[i] = dp[i - 1] + 1;
            used[x] = 1;
            ans[x] = {last[x], i};
          } else {
            last[x] = i;
            dp[i] = dp[i - 1];
          }
        } else
          dp[i] = dp[i - 1];
      }
    }
    for (int i = 0; i < n; i++)
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
  }
  void solve() {
    int T = 1;
    int TestCase = 0;
    while (T--) {
      TestCase += 1;
      sol();
    }
  }
  int main(int argc, char *argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    open_file();
    solve();
  }
  