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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  const long long mod = 1e9 + 7;
  const double PI = acos(-1.0);
  const double eps = 1e-8;
  long long powmod(long long a, long long b) {
    long long res = 1;
    a %= mod;
    assert(b >= 0);
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % mod;
      a = a * a % mod;
    }
    return res;
  }
  const int maxn = 2000005;
  int main() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n * k + 1), pos(n + 1), l(n + 1), r(n + 1);
    vector<pair<int, int> > segs;
    int mx = n / (k - 1) + (n % (k - 1) != 0);
    for (int i = 1; i <= (n * k); ++i) {
      cin >> a[i];
      if (pos[a[i]]) {
        segs.push_back(make_pair(pos[a[i]], i));
      }
      pos[a[i]] = i;
    }
    sort(segs.begin(), segs.end(),
         [](pair<int, int> x, pair<int, int> y) -> bool {
           return x.second < y.second;
         });
    vector<int> vis(n + 1), num(n * k + 1);
    for (auto it : segs) {
      if (vis[a[it.first]]) continue;
      bool flg = true;
      for (int i = it.first; i <= (it.second); ++i)
        if (num[i] + 1 > mx) {
          flg = false;
          break;
        }
      if (!flg) continue;
      int x = a[it.first];
      vis[x] = 1;
      l[x] = it.first, r[x] = it.second;
      for (int i = it.first; i <= it.second; ++i) num[i]++;
    }
    for (int i = 1; i <= (n); ++i) {
      cout << l[i] << " " << r[i] << endl;
    }
    return 0;
  }
  