#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  string sep;
  for (const T &x : v) os << sep << x, sep = " ";
  return os;
}
const int mxN = 1e5 + 1, oo = 1e9;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n * k);
  for (auto &i : a) cin >> i, --i;
  vector<pair<int, int> > ans(n, {-1, -1});
  int got = 0;
  while (got < n) {
    vector<int> prv(n, -1);
    int last = -1;
    for (int i = 0; i < n * k; ++i) {
      if (prv[a[i]] > last and ans[a[i]].first == -1) {
        ans[a[i]] = {prv[a[i]], i};
        last = i;
        got++;
      }
      prv[a[i]] = i;
    }
  }
  for (auto [aa, b] : ans) {
    cout << aa + 1 << ' ' << b + 1 << '\n';
  }
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  string sep;
  for (const T &x : v) os << sep << x, sep = " ";
  return os;
}
const int mxN = 1e5 + 1, oo = 1e9;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n * k);
  for (auto &i : a) cin >> i, --i;
  vector<pair<int, int> > ans(n, {-1, -1});
  int got = 0;
  while (got < n) {
    vector<int> prv(n, -1);
    int last = -1;
    for (int i = 0; i < n * k; ++i) {
      if (prv[a[i]] > last and ans[a[i]].first == -1) {
        ans[a[i]] = {prv[a[i]], i};
        last = i;
        got++;
      }
      prv[a[i]] = i;
    }
  }
  for (auto [aa, b] : ans) {
    cout << aa + 1 << ' ' << b + 1 << '\n';
  }
}

```