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 INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3fll;
  const long long M = 998244353;
  const long long maxn = 1.2e6 + 107;
  const double pi = acos(-1.0);
  const double eps = 0.0000000001;
  long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  template <typename T>
  inline void pr2(T x, unsigned long long k = 64) {
    unsigned long long i;
    for (i = 0; i < k; i++) fprintf(stderr, "%d", (int)((x >> i) & 1));
    putchar(' ');
  }
  template <typename T>
  inline void add_(T &A, int B, long long MOD = M) {
    A += B;
    (A >= MOD) && (A -= MOD);
  }
  template <typename T>
  inline void mul_(T &A, long long B, long long MOD = M) {
    A = (A * B) % MOD;
  }
  template <typename T>
  inline void mod_(T &A, long long MOD = M) {
    A %= MOD;
    A += MOD;
    A %= MOD;
  }
  template <typename T>
  inline void max_(T &A, T B) {
    (A < B) && (A = B);
  }
  template <typename T>
  inline void min_(T &A, T B) {
    (A > B) && (A = B);
  }
  template <typename T>
  inline T abs(T a) {
    return a > 0 ? a : -a;
  }
  inline long long powMM(long long a, long long b, long long mod = M) {
    long long ret = 1;
    for (; b; b >>= 1ll, a = a * a % mod)
      if (b & 1) ret = ret * a % mod;
    return ret;
  }
  int startTime;
  void startTimer() { startTime = clock(); }
  void printTimer() {
    fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
  }
  int A[maxn], R[maxn], id[maxn];
  pair<int, int> ans[maxn];
  set<int> Q;
  int main() {
    int T, _;
    T = 1;
    for (_ = 1; _ <= T; _++) {
      int n, i, K;
      scanf("%d%d", &n, &K);
      int maxCount = (n + K - 1 - 1) / (K - 1);
      for (i = 1; i <= n * K; i++) scanf("%d", &A[i]);
      for (i = n * K; i >= 1; i--) {
        R[i] = id[A[i]];
        id[A[i]] = i;
      }
      for (i = 1; i <= n * K; i++) {
        if (ans[A[i]].second == i) Q.erase(i);
        if (ans[A[i]].first == 0) ans[A[i]] = make_pair(i, R[i]), Q.insert(R[i]);
        while ((int)Q.size() > maxCount) {
          int x = *Q.rbegin();
          Q.erase(x);
          ans[A[x]] = make_pair(0, 0);
        }
      }
      for (i = 1; i <= n; i++) printf("%d %d\n", ans[i].first, ans[i].second);
    }
  }
  