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>
  #pragma GCC optimize("O3")
  #pragma GCC target("avx2")
  using namespace std;
  long long mod = 998244353;
  int mul(long long a, long long b) { return (a * b) % mod; }
  int binexp(long long a, long long b) {
    if (b == 0) return 1;
    long long half = binexp(a, b / 2);
    if ((b % 2) == 0)
      return mul(half, half);
    else
      return mul(a, mul(half, half));
  }
  void prop(int node, int tree[], int delta[]) {
    delta[node << 1] += delta[node];
    tree[node << 1] += delta[node];
    delta[node << 1 | 1] += delta[node];
    tree[node << 1 | 1] += delta[node];
    delta[node] = 0;
  }
  void update(int node, int l, int r, int start, int end, int tree[],
              int delta[]) {
    if (l > end || r < start || start > end) return;
    if (l <= start && end <= r) {
      tree[node] += 1;
      delta[node] += 1;
      return;
    }
    prop(node, tree, delta);
    int mid = start + (end - start) / 2;
    update(node << 1, l, r, start, mid, tree, delta);
    update(node << 1 | 1, l, r, mid + 1, end, tree, delta);
    tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
  }
  int query(int node, int l, int r, int start, int end, int tree[], int delta[]) {
    if (l > end || r < start || start > end) return INT_MIN;
    if (l <= start && end <= r) return tree[node];
    prop(node, tree, delta);
    int mid = start + (end - start) / 2;
    int p1 = query(node << 1, l, r, start, mid, tree, delta);
    int p2 = query(node << 1 | 1, l, r, mid + 1, end, tree, delta);
    tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
    return max(p1, p2);
  }
  bool cmp(pair<int, pair<int, int>> p1, pair<int, pair<int, int>> p2) {
    return (p1.first < p2.first) ||
           ((p1.first == p2.first && (p1.second.first > p2.second.first ||
                                      (p1.second.first == p2.second.first &&
                                       p1.second.second > p2.second.second))));
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    int t = 1;
    while (t--) {
      int n, k;
      cin >> n >> k;
      int arr[n * k];
      for (long long i = 0; i < n * k; ++i) {
        cin >> arr[i];
      }
      int tree[4 * n * k];
      memset(tree, 0, sizeof(tree));
      int last[n + 1];
      memset(last, -1, sizeof(last));
      int vis[n * k];
      memset(vis, false, sizeof(vis));
      int delta[4 * n * k];
      memset(delta, 0, sizeof(delta));
      vector<pair<int, pair<int, int>>> ans;
      for (long long i = 0; i < n * k; ++i) {
        if (last[arr[i]] == -1) {
          last[arr[i]] = i;
        } else if (vis[last[arr[i]]])
          continue;
        else {
          int size = (n + k - 2) / (k - 1);
          int temp = query(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
          if (temp < size) {
            update(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
            vis[i] = true;
            ans.push_back({arr[i], {last[arr[i]], i}});
          }
          last[arr[i]] = i;
        }
      }
      sort(ans.begin(), ans.end());
      for (long long i = 0; i < min(n, (int)ans.size()); ++i) {
        cout << ans[i].second.first + 1 << " " << ans[i].second.second + 1
             << endl;
      }
    }
  }
  