#include <bits/stdc++.h>
using namespace std;
const long long INF_LL = 4e18;
const int INF = 1e9 + 100;
const int MOD = 998244353;
const long double EPS = 1e-10;
const int BLOCK = 550;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
void open_file() {}
const int maxN = 2e6 + 100;
void sol() {
  int n, k;
  cin >> n >> k;
  vector<int> c(n * k);
  for (int &x : c) {
    cin >> x;
    x -= 1;
  }
  vector<int> used(n * k, 0);
  vector<int> last(n * k, -1);
  vector<int> dp(n * k, 0);
  vector<pair<int, int> > ans(n);
  for (int i = 0; i < n * k; i++) {
    int x = c[i];
    if (last[x] == -1) {
      last[x] = i;
      if (i == 0) continue;
      dp[i] = dp[i - 1];
    } else {
      if (used[x] == 0) {
        if (dp[i - 1] - dp[last[x]] < (n + k - 2) / (k - 1)) {
          dp[i] = dp[i - 1] + 1;
          used[x] = 1;
          ans[x] = {last[x], i};
        } else {
          last[x] = i;
          dp[i] = dp[i - 1];
        }
      } else
        dp[i] = dp[i - 1];
    }
  }
  for (int i = 0; i < n; i++)
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
}
void solve() {
  int T = 1;
  int TestCase = 0;
  while (T--) {
    TestCase += 1;
    sol();
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  open_file();
  solve();
}

### Prompt

Develop a solution in cpp 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;
const long long INF_LL = 4e18;
const int INF = 1e9 + 100;
const int MOD = 998244353;
const long double EPS = 1e-10;
const int BLOCK = 550;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
void open_file() {}
const int maxN = 2e6 + 100;
void sol() {
  int n, k;
  cin >> n >> k;
  vector<int> c(n * k);
  for (int &x : c) {
    cin >> x;
    x -= 1;
  }
  vector<int> used(n * k, 0);
  vector<int> last(n * k, -1);
  vector<int> dp(n * k, 0);
  vector<pair<int, int> > ans(n);
  for (int i = 0; i < n * k; i++) {
    int x = c[i];
    if (last[x] == -1) {
      last[x] = i;
      if (i == 0) continue;
      dp[i] = dp[i - 1];
    } else {
      if (used[x] == 0) {
        if (dp[i - 1] - dp[last[x]] < (n + k - 2) / (k - 1)) {
          dp[i] = dp[i - 1] + 1;
          used[x] = 1;
          ans[x] = {last[x], i};
        } else {
          last[x] = i;
          dp[i] = dp[i - 1];
        }
      } else
        dp[i] = dp[i - 1];
    }
  }
  for (int i = 0; i < n; i++)
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
}
void solve() {
  int T = 1;
  int TestCase = 0;
  while (T--) {
    TestCase += 1;
    sol();
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  open_file();
  solve();
}

```