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 int maxn = 105, pot = 16384;
  struct tournament {
    int t[pot * 2];
    int prop[pot * 2];
    void propagate(int x) {
      if (prop[x]) {
        t[x] += prop[x];
        if (x < pot) {
          prop[x * 2] += prop[x];
          prop[x * 2 + 1] += prop[x];
        }
        prop[x] = 0;
      }
    }
    void update(int x, int val) {
      t[x] += val;
      for (x /= 2; x > 0; x /= 2) {
        propagate(x * 2);
        propagate(x * 2 + 1);
        t[x] = max(t[x * 2], t[x * 2 + 1]);
      }
    }
    void update2(int L, int D, int x, int l, int d, int val) {
      propagate(x);
      if (L >= l && D <= d) {
        update(x, val);
        if (x < pot) {
          prop[x * 2] += val;
          prop[x * 2 + 1] += val;
        }
        return;
      }
      if ((L + D) / 2 >= l) {
        update2(L, (L + D) / 2, x * 2, l, d, val);
      }
      if ((L + D) / 2 + 1 <= d) {
        update2((L + D) / 2 + 1, D, x * 2 + 1, l, d, val);
      }
    }
    int query(int L, int D, int x, int l, int d) {
      propagate(x);
      if (L >= l && D <= d) {
        return t[x];
      }
      int lijeva = 0, desna = 0;
      if ((L + D) / 2 >= l) {
        lijeva = query(L, (L + D) / 2, x * 2, l, d);
      }
      if ((L + D) / 2 + 1 <= d) {
        desna = query((L + D) / 2 + 1, D, x * 2 + 1, l, d);
      }
      return max(lijeva, desna);
    }
  };
  tournament T;
  vector<pair<int, int> > v;
  int a[maxn * maxn];
  int pos[maxn];
  int ind[maxn];
  vector<int> boja[maxn];
  bool cmp(int x, int y) { return boja[x][ind[x]] > boja[y][ind[y]]; }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n * k; i++) {
      cin >> a[i];
      a[i]--;
      boja[a[i]].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      pos[i] = i;
      ind[i] = 1;
      T.update2(0, pot - 1, 1, boja[i][0], boja[i][1], 1);
    }
    bool p = 1;
    int maksi = (n + k - 2) / (k - 1);
    int x;
    while (p) {
      p = 0;
      sort(pos, pos + n, cmp);
      for (int i = 0; i < n; i++) {
        x = pos[i];
        if (T.query(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]]) >
            maksi) {
          p = 1;
          T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], -1);
          ind[x]++;
          T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], 1);
        }
      }
    }
    for (int i = 0; i < n; i++) {
      cout << boja[i][ind[i] - 1] + 1 << ' ' << boja[i][ind[i]] + 1 << '\n';
    }
    return 0;
  }
  