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")
  using namespace std;
  using ll = long long int;
  using ld = long double;
  using P = pair<ll, ll>;
  using Graph = vector<vector<ll>>;
  struct edge {
    ll to;
    ll cost;
  };
  using graph = vector<vector<edge>>;
  const ll INF = 1e18;
  inline void chmax(ll& a, ll b) { a = max(a, b); }
  inline void chmin(ll& a, ll b) { a = min(a, b); }
  long double pi = acos(-1);
  ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
  ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
  ll dx[4]{1, 0, -1, 0};
  ll dy[4]{0, 1, 0, -1};
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n, k;
    cin >> n >> k;
    ll m = n * k;
    vector<ll> A(m);
    for (ll i = 0; i < (m); ++i) cin >> A[i];
    vector<vector<ll>> B(n);
    for (ll i = 0; i < (m); ++i) B[A[i] - 1].push_back(i);
    vector<vector<P>> unti(n);
    for (ll i = 0; i < (n); ++i)
      for (ll j = 0; j < (k - 1); ++j) {
        ll x = B[i][j];
        ll y = B[i][j + 1];
        unti.push_back({make_pair(x, y)});
      }
    vector<ll> seen(n);
    vector<P> ans(n);
    ll cnt = 0;
    bool ok = 0;
    for (ll i = 0; i < (n); ++i) {
      ll now = -1;
      while (1) {
        ll ma = INF;
        ll next = -1;
        ll soeji = -1;
        for (ll j = 0; j < (n); ++j) {
          if (seen[j]) continue;
          ll y = upper_bound(B[j].begin(), B[j].end(), now) - B[j].begin();
          if (y == k - 1 || y == k) continue;
          ll x = B[j][y + 1];
          if (x < ma) {
            ma = x;
            next = j;
            soeji = y;
          }
        }
        if (ma == INF) break;
        seen[next] = true;
        ans[next].first = B[next][soeji];
        ans[next].second = B[next][soeji + 1];
        now = ma;
        cnt++;
        if (cnt == n) {
          ok = 1;
          break;
        }
      }
      if (ok) break;
    }
    for (ll i = 0; i < (n); ++i)
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
    return 0;
  }
  