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;
  array<long long int, 2> ans[103];
  long long int prev_pos[103];
  bool flg[103];
  long long int st[80303];
  long long int lazy[80303];
  long long int a[10003];
  void do_lazy(long long int node, long long int l, long long int r) {
    st[node] += lazy[node];
    if (l != r) {
      lazy[2ll * node + 1ll] += lazy[node];
      lazy[2ll * node + 2ll] += lazy[node];
    }
    lazy[node] = 0;
  }
  void upd(long long int node, long long int l, long long int r, long long int ql,
           long long int qr, long long int val) {
    do_lazy(node, l, r);
    if (l > r || l > qr || r < ql) {
      return;
    }
    if (l >= ql && r <= qr) {
      lazy[node] += val;
      do_lazy(node, l, r);
      return;
    }
    long long int mid = (l + r) / 2;
    upd(2ll * node + 1ll, l, mid, ql, qr, val);
    upd(2ll * node + 2ll, mid + 1ll, r, ql, qr, val);
    st[node] = max(st[2ll * node + 1ll], st[2ll * node + 2ll]);
  }
  long long int query(long long int node, long long int l, long long int r,
                      long long int ql, long long int qr) {
    do_lazy(node, l, r);
    if (l > r || l > qr || r < ql) {
      return 0;
    }
    if (l >= ql && r <= qr) {
      return st[node];
    }
    long long int mid = (l + r) / 2;
    return max(query(2ll * node + 1ll, l, mid, ql, qr),
               query(2ll * node + 2ll, mid + 1ll, r, ql, qr));
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int n, k;
    cin >> n >> k;
    for (long long int i = 1ll; i <= n * k; i++) {
      cin >> a[i];
    }
    for (long long int i = 1ll; i <= n * k; i++) {
      if (flg[a[i]]) {
        continue;
      }
      if (prev_pos[a[i]] == 0) {
        prev_pos[a[i]] = i;
        continue;
      }
      long long int val = query(1, 1, n * k, prev_pos[a[i]], i - 1ll);
      if (val >= ((n + k - 2ll) / (k - 1ll))) {
        prev_pos[a[i]] = i;
      } else {
        upd(1, 1, n * k, prev_pos[a[i]], i, 1);
        flg[a[i]] = 1;
        ans[a[i]] = {prev_pos[a[i]], i};
      }
    }
    for (long long int i = 1ll; i <= n; i++) {
      cout << ans[i][0] << " " << ans[i][1] << "\n";
    }
  }
  