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;
  inline int getint() {
    int n;
    scanf("%d", &n);
    return n;
  }
  vector<int> a[105];
  vector<pair<int, int> > b;
  int sg[100005], add[100005];
  bool lazy[100005];
  void push_down(int rt) {
    lazy[rt << 1] = lazy[rt << 1 | 1] = true;
    add[rt << 1] += add[rt];
    add[rt << 1 | 1] += add[rt];
    sg[rt << 1] += add[rt];
    sg[rt << 1 | 1] += add[rt];
    add[rt] = 0;
    lazy[rt] = false;
  }
  void push_up(int rt) { sg[rt] = max(sg[rt << 1], sg[rt << 1 | 1]); }
  int get_max(int L, int R, int rt, int l, int r) {
    if (L <= l && r <= R) return sg[rt];
    if (lazy[rt]) push_down(rt);
    int ans = 0;
    int mid = (l + r) >> 1;
    if (L <= mid) ans = max(ans, get_max(L, R, rt << 1, l, mid));
    if (R > mid) ans = max(ans, get_max(L, R, rt << 1 | 1, mid + 1, r));
    return ans;
  }
  void add_one(int L, int R, int rt, int l, int r) {
    if (L <= l && r <= R) {
      lazy[rt] = true;
      add[rt] += 1;
      sg[rt] += 1;
      return;
    }
    if (lazy[rt]) push_down(rt);
    int mid = (l + r) >> 1;
    if (L <= mid) add_one(L, R, rt << 1, l, mid);
    if (R > mid) add_one(L, R, rt << 1 | 1, mid + 1, r);
    push_up(rt);
  }
  bool cmp(const pair<int, int> &x, const pair<int, int> &y) {
    return x.second < y.second;
  }
  bool used[105];
  int record[10005];
  vector<pair<int, int> > ans;
  int main() {
    int n = getint(), K = getint();
    for (int i = 1; i <= n * K; i++) {
      int c = getint();
      record[i] = c;
      a[c].emplace_back(i);
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < a[i].size(); j++) {
        for (int k = j + 1; k < a[i].size(); k++) {
          b.emplace_back(pair<int, int>(a[i][j], a[i][k]));
        }
      }
    }
    sort(b.begin(), b.end(), cmp);
    int M = int(n + K - 2) / (K - 1);
    int cnt = n;
    ans.resize(n + 1);
    for (int i = 0; i < b.size(); i++) {
      if (cnt == 0) break;
      int col = record[b[i].first];
      if (used[col]) continue;
      if (get_max(b[i].first, b[i].second, 1, 1, n * K) < M) {
        ans[col] = pair<int, int>(b[i].first, b[i].second);
        cnt--;
        used[col] = true;
        add_one(b[i].first, b[i].second, 1, 1, n * K);
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d %d\n", ans[i].first, ans[i].second);
    }
    return 0;
  }
  