#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int a[100005];
vector<int> c[105];
int ans[105][2];
int idx[105], vidx[105], vis[105];
void go(int gn, int kk) {
  if (gn == 0 || kk < 2) {
    return;
  }
  for (int i = (1); i <= (gn); ++i) {
    vis[i] = 0;
  }
  while (1) {
    int mn = 1e9, minidx;
    for (int i = (1); i <= (gn); ++i) {
      if (c[idx[i]][vidx[i]] < mn) {
        mn = c[idx[i]][vidx[i]];
        minidx = i;
      }
    }
    if (vis[minidx] != 0) {
      ans[idx[minidx]][0] = vis[minidx];
      ans[idx[minidx]][1] = mn;
      idx[minidx] = idx[gn];
      vidx[minidx] = vidx[gn];
      gn--;
      go(gn, kk - 1);
      return;
    } else {
      vis[minidx] = mn;
      vidx[minidx]++;
    }
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) {
    c[i].clear();
  }
  for (int i = (1); i <= (n * k); ++i) {
    cin >> a[i];
    c[a[i]].push_back(i);
  }
  int lay = n / (k - 1) + ((n % (k - 1)) == 0 ? 0 : 1);
  for (int i = (1); i <= (lay); ++i) {
    int l = 0;
    for (int j = (1); j <= (k - 1); ++j) {
      if ((i - 1) * (k - 1) + j > n) {
        break;
      }
      l++;
      idx[j] = (i - 1) * (k - 1) + j;
      vidx[j] = 0;
    }
    go(l, k);
  }
  for (int i = (1); i <= (n); ++i) {
    cout << ans[i][0] << " " << ans[i][1] << "\n";
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
using ll = long long;
int a[100005];
vector<int> c[105];
int ans[105][2];
int idx[105], vidx[105], vis[105];
void go(int gn, int kk) {
  if (gn == 0 || kk < 2) {
    return;
  }
  for (int i = (1); i <= (gn); ++i) {
    vis[i] = 0;
  }
  while (1) {
    int mn = 1e9, minidx;
    for (int i = (1); i <= (gn); ++i) {
      if (c[idx[i]][vidx[i]] < mn) {
        mn = c[idx[i]][vidx[i]];
        minidx = i;
      }
    }
    if (vis[minidx] != 0) {
      ans[idx[minidx]][0] = vis[minidx];
      ans[idx[minidx]][1] = mn;
      idx[minidx] = idx[gn];
      vidx[minidx] = vidx[gn];
      gn--;
      go(gn, kk - 1);
      return;
    } else {
      vis[minidx] = mn;
      vidx[minidx]++;
    }
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) {
    c[i].clear();
  }
  for (int i = (1); i <= (n * k); ++i) {
    cin >> a[i];
    c[a[i]].push_back(i);
  }
  int lay = n / (k - 1) + ((n % (k - 1)) == 0 ? 0 : 1);
  for (int i = (1); i <= (lay); ++i) {
    int l = 0;
    for (int j = (1); j <= (k - 1); ++j) {
      if ((i - 1) * (k - 1) + j > n) {
        break;
      }
      l++;
      idx[j] = (i - 1) * (k - 1) + j;
      vidx[j] = 0;
    }
    go(l, k);
  }
  for (int i = (1); i <= (n); ++i) {
    cout << ans[i][0] << " " << ans[i][1] << "\n";
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}

```