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>
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int n, k;
  cin >> n >> k;
  vector<int> a(n * k);
  vector<vi> occ(n);
  for (int i = 0; i < n * k; ++i) {
    cin >> a[i];
    --a[i];
    occ[a[i]].push_back(i);
  }
  int goal = (n + k - 2) / (k - 1);
  int l = 0;
  vi freq(n, 0);
  vi done(n, 0);
  vi ready;
  vector<pii> ans(n, {-1, -1});
  for (int r = 0; r < n * k; ++r) {
    if (!done[a[r]] && ++freq[a[r]] == 2) {
      ready.push_back(a[r]);
    }
    if (ready.size() == goal) {
      for (int x : ready) {
        auto it = upper_bound(begin(occ[x]), end(occ[x]), r);
        assert(it - begin(occ[x]) >= 2);
        ans[x].second = *prev(it);
        ans[x].first = *prev(prev(it));
        assert(l <= ans[x].first && ans[x].first < ans[x].second &&
               ans[x].second <= r);
        done[x] = true;
      }
      freq.assign(n, 0);
      ready.clear();
      l = r + 1;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (!done[i]) {
      int m = (int)(occ[i]).size();
      ans[i].first = occ[i][m - 2];
      ans[i].second = occ[i][m - 1];
    }
  }
  for (int i = 0; i < n; ++i) {
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
  }
  return 0;
}
