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 = 2e5 + 5;
  const int M = 2e3 + 5;
  const long long mod = 998244353;
  const long double ex = 1e-7;
  long long m, n, k, a[N], ans, tong, idx, mn, c[N], d[N], t, b[N], lab[N], id[N],
      ok[N], pos[M][M], l[M], r[M];
  vector<pair<long long, long long> > adj[N];
  vector<pair<long long, long long> > kq;
  long long pw(long long k, long long n) {
    long long total = 1;
    for (; n; n >>= 1) {
      if (n & 1) total = total * k % mod;
      k = k * k % mod;
    }
    return total;
  }
  void add(long long id, long long x, long long fe[]) {
    for (; id <= m; id += id & -id) fe[id] = min(fe[id], x);
  }
  long long lwr(long long x, vector<long long> kq) {
    long long i = lower_bound(kq.begin(), kq.end(), x) - kq.begin();
    if (i == kq.size() || kq[i] != x) return -1;
    return i;
  }
  long long C(long long u, long long v) {
    if (u > v) return 0;
    return a[v] * b[u] % mod * b[v - u] % mod;
  }
  void fixv(vector<pair<long long, long long> >& v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
  }
  long long findp(long long u) { return lab[u] < 0 ? u : lab[u] = findp(lab[u]); }
  bool cmp(const long long& x, const long long& y) { return a[x] < a[y]; }
  void sol() {
    cin >> n >> k;
    m = (n + k - 2) / (k - 1);
    for (int i = 1; i <= n * k; i++) {
      cin >> a[i];
      ++b[a[i]];
      pos[a[i]][b[a[i]]] = i;
      adj[b[a[i]]].push_back({i, a[i]});
    }
    for (int i = 1; i <= n; i++) id[i] = i;
    for (int i = 2; i <= k; i++) {
      sort(adj[i].begin(), adj[i].end());
      long long cnt = 0;
      for (pair<long long, long long> x : adj[i]) {
        if (d[x.second]) continue;
        d[x.second] = 1;
        r[x.second] = x.first;
        l[x.second] = pos[x.second][i - 1];
        ++cnt;
        if (cnt == m) break;
      }
    }
    for (int i = 1; i <= n; i++) cout << l[i] << " " << r[i] << '\n';
  }
  int main() {
    cin.tie(0);
    cout.tie(0);
    ios_base::sync_with_stdio(0);
    srand((int)time(0));
    int ntest = 1;
    while (ntest-- > 0) sol();
    return 0;
  }
  