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;
  void JIZZ(string output = "") {
    cout << output;
    exit(0);
  }
  const long double PI = 3.14159265358979323846264338327950288;
  const long double eps = 1e-10;
  const long long mod = 1e9 + 7;
  struct node {
    node *l, *r;
    int mx, tag;
    node() : l(0), r(0), mx(0), tag(0) {}
    void pull() { mx = max(l->mx, r->mx); }
    void push() {
      if (tag) l->addtag(tag), r->addtag(tag), tag = 0;
    }
    void addtag(int t) { mx += t, tag += t; }
  } * root;
  void build(node *now, int l, int r) {
    if (l == r) {
      return;
    }
    build(now->l = new node(), l, ((l + r) >> 1));
    build(now->r = new node(), ((l + r) >> 1) + 1, r);
  }
  void modify(node *now, int l, int r, int ml, int mr, int v) {
    if (r < ml || mr < l) return;
    if (ml <= l && r <= mr) {
      now->addtag(v);
      return;
    }
    now->push();
    modify(now->l, l, ((l + r) >> 1), ml, mr, v);
    modify(now->r, ((l + r) >> 1) + 1, r, ml, mr, v);
    now->pull();
  }
  int query(node *now, int l, int r, int ql, int qr) {
    if (r < ql || qr < l) return -1;
    if (ql <= l && r <= qr) return now->mx;
    now->push();
    return max(query(now->l, l, ((l + r) >> 1), ql, qr),
               query(now->r, ((l + r) >> 1) + 1, r, ql, qr));
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int n, k;
    cin >> n >> k;
    int most = n / (k - 1) + !!(n % (k - 1));
    int nn = n * k;
    vector<int> c(nn);
    for (int i = 0; i < nn; ++i) cin >> c[i], --c[i];
    vector<int> cpre(n, -1);
    vector<int> pre(nn);
    for (int i = 0; i < nn; ++i) {
      pre[i] = cpre[c[i]];
      cpre[c[i]] = i;
    }
    build(root = new node(), 0, nn - 1);
    vector<pair<int, int>> pos(n, make_pair(-1, -1));
    for (int i = 1; i < nn; ++i) {
      if (pos[c[i]] == make_pair(-1, -1)) {
        if (pre[i] != -1) {
          if (query(root, 0, nn - 1, pre[i], i) < most) {
            modify(root, 0, nn - 1, pre[i], i, 1);
            pos[c[i]] = make_pair(pre[i], i);
          }
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      assert(pos[i] != make_pair(-1, -1));
      cout << pos[i].first + 1 << ' ' << pos[i].second + 1 << '\n';
    }
  }
  