#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n * k);
  for (long long& i : v) {
    cin >> i;
    --i;
  }
  vector<vector<long long>> pos(n);
  for (long long i = 0; i < n * k; ++i) pos[v[i]].push_back(i);
  vector<pair<long long, long long>> ans(n), rng;
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < k - 1; ++j) {
      rng.push_back({pos[i][j], pos[i][j + 1]});
    }
  }
  sort(rng.begin(), rng.end(),
       [&](pair<long long, long long> a, pair<long long, long long> b) {
         return a.second < b.second;
       });
  pair<long long, long long> zero = {0, 0};
  long long bound = (n + k - 2) / (k - 1);
  vector<long long> full(n * k);
  for (long long i = 0; i < rng.size(); ++i) {
    long long col = v[rng[i].first];
    if (ans[col] != zero) continue;
    long long y = 1;
    for (long long x = rng[i].first; x <= rng[i].second; ++x)
      y &= full[x] < bound;
    if (!y) continue;
    for (long long x = rng[i].first; x <= rng[i].second; ++x) ++full[x];
    ans[col] = rng[i];
  }
  assert(ans.back() != zero);
  for (auto [a, b] : ans) cout << a + 1 << " " << b + 1 << endl;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n * k);
  for (long long& i : v) {
    cin >> i;
    --i;
  }
  vector<vector<long long>> pos(n);
  for (long long i = 0; i < n * k; ++i) pos[v[i]].push_back(i);
  vector<pair<long long, long long>> ans(n), rng;
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < k - 1; ++j) {
      rng.push_back({pos[i][j], pos[i][j + 1]});
    }
  }
  sort(rng.begin(), rng.end(),
       [&](pair<long long, long long> a, pair<long long, long long> b) {
         return a.second < b.second;
       });
  pair<long long, long long> zero = {0, 0};
  long long bound = (n + k - 2) / (k - 1);
  vector<long long> full(n * k);
  for (long long i = 0; i < rng.size(); ++i) {
    long long col = v[rng[i].first];
    if (ans[col] != zero) continue;
    long long y = 1;
    for (long long x = rng[i].first; x <= rng[i].second; ++x)
      y &= full[x] < bound;
    if (!y) continue;
    for (long long x = rng[i].first; x <= rng[i].second; ++x) ++full[x];
    ans[col] = rng[i];
  }
  assert(ans.back() != zero);
  for (auto [a, b] : ans) cout << a + 1 << " " << b + 1 << endl;
}

```