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 N = 100;
  inline int read() {
    int ret = 0, op = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') op = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      ret = ret * 10 + ch - '0';
      ch = getchar();
    }
    return ret * op;
  }
  inline long long readll() {
    long long ret = 0, op = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') op = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      ret = ret * 10 + ch - '0';
      ch = getchar();
    }
    return ret * op;
  }
  void write(int x) {
    if (x < 0) {
      putchar('-');
      x = -x;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  struct node {
    int l, r, col;
  } ans[N + 15], c[N * N + 15];
  inline bool cmp(node e, node f) {
    if (e.r == f.r) return e.l < f.l;
    return e.r < f.r;
  }
  inline bool cmp1(node e, node f) { return e.col < f.col; }
  int a[N * N + 15], pre[N * N + 15];
  int vis[N + 15];
  int cnt[N * N + 15];
  int main() {
    int n = read();
    int k = read();
    int mxcnt = n / (k - 1) + (n % (k - 1) != 0);
    int tot = 0;
    for (register int i = 1; i <= n * k; i++) {
      a[i] = read();
      if (pre[a[i]]) {
        c[++tot] = (node){pre[a[i]], i, a[i]};
      }
      pre[a[i]] = i;
    }
    sort(c + 1, c + tot + 1, cmp);
    int sum = 0;
    for (register int i = 1; i <= tot; i++) {
      if (vis[c[i].col]) continue;
      bool flag = 1;
      for (register int j = c[i].l; j <= c[i].r; j++) {
        if (cnt[j] == mxcnt) {
          flag = 0;
          break;
        }
      }
      if (!flag) continue;
      ans[++sum] = c[i];
      vis[c[i].col] = 1;
      for (register int j = c[i].l; j <= c[i].r; j++) {
        cnt[j]++;
      }
    }
    sort(ans + 1, ans + sum + 1, cmp1);
    for (register int i = 1; i <= sum; i++) {
      printf("%d %d\n", ans[i].l, ans[i].r);
    }
    return 0;
  }
  