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;
  long long n, m, k;
  long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      s = s * 10 + ch - '0';
      ch = getchar();
    }
    return s * f;
  }
  const long long maxn = 3e5 + 10;
  long long a[maxn];
  long long pre[maxn];
  long long f[maxn];
  long long x[maxn], y[maxn];
  long long in(long long l, long long r, long long x) { return x >= l && x <= r; }
  void clr(long long l, long long r) {
    for (long long i = l, iend = r; i <= iend; ++i) pre[i] = 0;
  }
  signed main() {
    n = read(), k = read();
    long long siz = n * k;
    for (long long i = 1, iend = siz; i <= iend; ++i) a[i] = read();
    long long u = k - 1;
    long long l = 1, r = u;
    while (l <= n) {
      if (r > n) r = n;
      for (long long i = 1, iend = siz; i <= iend; ++i) {
        long long v = a[i];
        if (!in(l, r, v)) continue;
        if (f[v]) continue;
        if (!pre[v]) {
          pre[v] = i;
          continue;
        }
        x[v] = pre[v];
        y[v] = i;
        f[v] = 1;
        clr(l, r);
      }
      l += u;
      r += u;
    }
    for (long long i = 1, iend = n; i <= iend; ++i) {
      cout << x[i] << " " << y[i] << '\n';
    }
  }
  