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;
const long long MAXN = 16000001;
const long long inf = 1e18;
using ll = long long;
void fast_stream() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long maxn = 105;
const long long maxnk = 10005;
vector<ll> a(maxnk, 0);
vector<ll> used(maxn, 0);
vector<ll> x(maxn), y(maxn);
void solve() {
  ll n, k;
  cin >> n >> k;
  for (long long i = 1; i <= n * k; ++i) cin >> a[i];
  long long ch = 0;
  for (long long i = 1; i <= n * k; i++) {
    if (used[a[i]] == 0) {
      used[a[i]] = 1;
      x[a[i]] = i;
    } else if (used[a[i]] == 1) {
      used[a[i]] = 2;
      y[a[i]] = i;
      ++ch;
    }
    if (ch == (n + (k - 2)) / (k - 1)) {
      for (long long j = 1; j <= n; j++)
        if (used[j] == 1) used[j] = 0;
      ch = 0;
    }
  }
  for (long long i = 1; i <= n; ++i) {
    cout << x[i] << ' ' << y[i] << '\n';
  }
  return;
}
signed main() {
  fast_stream();
  ll t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
