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 MOD = 1e9 + 7;
  const long double PI = 3.141592653589793238462643383279502884197;
  long long fac[1] = {1}, inv[1] = {1};
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long mp(long long a, long long b) {
    long long ret = 1;
    while (b) {
      if (b & 1) ret = ret * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return ret;
  }
  long long cmb(long long r, long long c) {
    return (c > r || c < 0) ? 0 : fac[r] * inv[c] % MOD * inv[r - c] % MOD;
  }
  int vec[10000];
  int ok[100];
  int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n * k; i++) {
      scanf("%d", &vec[i]);
      vec[i]--;
    }
    vector<pair<int, int> > ans;
    ans.resize(n);
    map<int, int> ma;
    while (1) {
      int ff = 0;
      for (int i = 0; i < n; i++)
        if (ok[i] == 0) ff = 1;
      if (ff == 0) break;
      ma.clear();
      for (int i = 0; i < n * k; i++) {
        if (ok[vec[i]])
          continue;
        else {
          if (ma.find(vec[i]) == ma.end())
            ma[vec[i]] = i;
          else {
            ans[vec[i]] = {ma[vec[i]], i};
            ok[vec[i]] = 1;
            ma.clear();
          }
        }
      }
    }
    for (pair<int, int> i : ans) printf("%d %d\n", i.first + 1, i.second + 1);
  }
  