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("-O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  mt19937_64 rnd(time(0));
  long long MOD = 998244353;
  const long double E = 1e-10;
  long long n, m, k, T, nw[100001], a[100001];
  pair<long long, long long> ans[100001];
  long long binpow(long long h, long long r, long long md) {
    long long l = 1;
    while (r) {
      if (r & 1) l = (l * h) % md;
      h = (h * h) % md;
      r /= 2;
    }
    return l;
  }
  long long read() {
    char c = getchar();
    while (!isdigit(c)) {
      c = getchar();
    }
    long long res = 0;
    while (isdigit(c)) {
      res = res * 10 + c - '0';
      c = getchar();
    }
    return res;
  }
  long long gcd(long long x, long long y) {
    while (x && y) {
      if (x < y) swap(x, y);
      x %= y;
    }
    return x + y;
  }
  bool tt[1001];
  vector<long long> c[1001];
  vector<long long> v[1001];
  void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) c[i].clear(), tt[i] = 0;
    for (int i = 1; i <= n * k; i++) {
      cin >> a[i];
      c[a[i]].push_back(i);
    }
    vector<long long> z;
    for (int i = 1; i <= n; i++) {
      z.push_back(i);
    }
    while (1) {
      long long y = 0;
      for (int i = 1; i <= n; i++)
        if (!tt[i]) nw[i] = 0, y = 1;
      if (!y) break;
      long long p = 0;
      while (1) {
        long long mn = 1e18, pp = 0, u = 0;
        for (int i = 1; i <= n; i++)
          if (!tt[i]) {
            while (nw[i] < c[i].size() - 1 && c[i][nw[i]] <= p) nw[i]++;
            if (nw[i] < c[i].size() - 1 && c[i][nw[i] + 1] < mn)
              mn = c[i][nw[i] + 1], u = c[i][nw[i]], pp = i;
          }
        if (pp == 0) break;
        tt[pp] = 1;
        ans[pp] = make_pair(u, mn);
        p = mn;
      }
    }
    for (int i = 1; i <= n; i++) {
      cout << ans[i].first << " " << ans[i].second << "\n";
    }
  }
  int main() {
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  