#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class S>
ostream &operator<<(ostream &o, const pair<T, S> &p) {
  return o << '(' << p.first << ", " << p.second << ')';
}
template <template <class, class...> class T, class... A>
typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
    ostream &o, T<A...> V) {
  o << '[';
  for (auto a : V) o << a << ", ";
  return o << ']';
}
long long int a[100010], l[100010], r[100010], pos[100010];
bool used[100010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n;
  cin >> n;
  long long int k;
  cin >> k;
  for (long long int i = 1; i < (n * k + 1); ++i) cin >> a[i];
  for (long long int t = 0; t < (n); ++t) {
    fill_n(pos, n + 1, -1);
    for (long long int i = 1; i < (n * k + 1); ++i) {
      if (~pos[a[i]] && !used[a[i]]) {
        l[a[i]] = pos[a[i]];
        r[a[i]] = i;
        used[a[i]] = true;
        fill_n(pos, n + 1, -1);
      } else
        pos[a[i]] = i;
    }
  }
  for (long long int i = 1; i < (n + 1); ++i)
    cout << l[i] << ' ' << r[i] << '\n';
}

### Prompt

Your task is to create a cpp solution to the following problem:

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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T, class S>
ostream &operator<<(ostream &o, const pair<T, S> &p) {
  return o << '(' << p.first << ", " << p.second << ')';
}
template <template <class, class...> class T, class... A>
typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
    ostream &o, T<A...> V) {
  o << '[';
  for (auto a : V) o << a << ", ";
  return o << ']';
}
long long int a[100010], l[100010], r[100010], pos[100010];
bool used[100010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n;
  cin >> n;
  long long int k;
  cin >> k;
  for (long long int i = 1; i < (n * k + 1); ++i) cin >> a[i];
  for (long long int t = 0; t < (n); ++t) {
    fill_n(pos, n + 1, -1);
    for (long long int i = 1; i < (n * k + 1); ++i) {
      if (~pos[a[i]] && !used[a[i]]) {
        l[a[i]] = pos[a[i]];
        r[a[i]] = i;
        used[a[i]] = true;
        fill_n(pos, n + 1, -1);
      } else
        pos[a[i]] = i;
    }
  }
  for (long long int i = 1; i < (n + 1); ++i)
    cout << l[i] << ' ' << r[i] << '\n';
}

```