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;
  class SegTree {
    int leftMost, rightMost, toProp, mxVal;
    SegTree *lChild, *rChild;
  
   public:
    SegTree(int left, int right)
        : leftMost(left), rightMost(right), toProp(0), mxVal(0) {
      if (left != right) {
        int mid = (left + right) / 2;
        lChild = new SegTree(left, mid);
        rChild = new SegTree(mid + 1, right);
      } else {
        lChild = rChild = NULL;
      }
    }
    int getMax() { return mxVal + toProp; }
    void update(int left, int right, int delta = 1) {
      if (left > rightMost || right < leftMost)
        return;
      else if (left <= leftMost && right >= rightMost) {
        toProp += delta;
        return;
      } else {
        prop();
        lChild->update(left, right, delta);
        rChild->update(left, right, delta);
        recalc();
      }
    }
    int query(int left, int right) {
      if (left > rightMost || right < leftMost)
        return -1;
      else if (left <= leftMost && right >= rightMost) {
        return getMax();
      } else {
        prop();
        return max(lChild->query(left, right), rChild->query(left, right));
      }
    }
    void prop() {
      if (lChild) {
        lChild->toProp += toProp;
        rChild->toProp += toProp;
        toProp = 0;
      }
      recalc();
    }
    void recalc() {
      if (lChild) mxVal = max(lChild->getMax(), rChild->getMax());
    }
  };
  int main() {
    int n, k;
    cin >> n >> k;
    vector<int> inp(n * k);
    for (auto& u : inp) {
      cin >> u;
      --u;
    }
    SegTree st(0, n * k - 1);
    vector<int> lastSeen(n, -1);
    vector<bool> done(n, false);
    vector<pair<int, int>> ans(n);
    int maxAllowed = (n + k - 2) / (k - 1);
    for (int i = 0; i < n * k; i++) {
      if (lastSeen[inp[i]] == -1 || done[inp[i]]) {
        lastSeen[inp[i]] = i;
      } else {
        int chk = lastSeen[inp[i]];
        int getMost = st.query(chk, i);
        if (getMost < maxAllowed) {
          st.update(chk, i);
          done[inp[i]] = true;
          ans[inp[i]] = {chk + 1, i + 1};
        } else {
          lastSeen[inp[i]] = i;
        }
      }
    }
    for (auto u : ans) cout << u.first << " " << u.second << '\n';
    return 0;
  }
  