#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-8;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int maxn = 2000005;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n * k + 1), pos(n + 1), l(n + 1), r(n + 1);
  vector<pair<int, int> > segs;
  int mx = n / (k - 1) + (n % (k - 1) != 0);
  for (int i = 1; i <= (n * k); ++i) {
    cin >> a[i];
    if (pos[a[i]]) {
      segs.push_back(make_pair(pos[a[i]], i));
    }
    pos[a[i]] = i;
  }
  sort(segs.begin(), segs.end(),
       [](pair<int, int> x, pair<int, int> y) -> bool {
         return x.second < y.second;
       });
  vector<int> vis(n + 1), num(n * k + 1);
  for (auto it : segs) {
    if (vis[a[it.first]]) continue;
    bool flg = true;
    for (int i = it.first; i <= (it.second); ++i)
      if (num[i] + 1 > mx) {
        flg = false;
        break;
      }
    if (!flg) continue;
    int x = a[it.first];
    vis[x] = 1;
    l[x] = it.first, r[x] = it.second;
    for (int i = it.first; i <= it.second; ++i) num[i]++;
  }
  for (int i = 1; i <= (n); ++i) {
    cout << l[i] << " " << r[i] << endl;
  }
  return 0;
}

### 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 optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long mod = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-8;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
const int maxn = 2000005;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n * k + 1), pos(n + 1), l(n + 1), r(n + 1);
  vector<pair<int, int> > segs;
  int mx = n / (k - 1) + (n % (k - 1) != 0);
  for (int i = 1; i <= (n * k); ++i) {
    cin >> a[i];
    if (pos[a[i]]) {
      segs.push_back(make_pair(pos[a[i]], i));
    }
    pos[a[i]] = i;
  }
  sort(segs.begin(), segs.end(),
       [](pair<int, int> x, pair<int, int> y) -> bool {
         return x.second < y.second;
       });
  vector<int> vis(n + 1), num(n * k + 1);
  for (auto it : segs) {
    if (vis[a[it.first]]) continue;
    bool flg = true;
    for (int i = it.first; i <= (it.second); ++i)
      if (num[i] + 1 > mx) {
        flg = false;
        break;
      }
    if (!flg) continue;
    int x = a[it.first];
    vis[x] = 1;
    l[x] = it.first, r[x] = it.second;
    for (int i = it.first; i <= it.second; ++i) num[i]++;
  }
  for (int i = 1; i <= (n); ++i) {
    cout << l[i] << " " << r[i] << endl;
  }
  return 0;
}

```