#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long n, k, t, m, q, flag = 0;
long long power(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;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    long long a[n * k];
    for (int i = 0; i < n * k; i++) cin >> a[i];
    pair<long long, long long> ans[n + 1];
    map<long long, long long> last;
    map<long long, long long> occ;
    long long temp = 2;
    long long val = n / (k - 1);
    if (n % (k - 1) != 0) val++;
    long long curr = 0;
    for (int i = 0; i < n * k; i++) {
      if (last[a[i]] == -1) continue;
      occ[a[i]]++;
      if (occ[a[i]] == temp) {
        ans[a[i]] = make_pair(last[a[i]], i);
        last[a[i]] = -1;
        curr++;
        if (curr % val == 0) temp++;
      } else {
        last[a[i]] = i;
      }
    }
    for (int i = 1; i <= n; i++)
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long mod = 1000000007;
long long n, k, t, m, q, flag = 0;
long long power(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;
}
long long min(long long a, long long b) {
  if (a > b)
    return b;
  else
    return a;
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    cin >> n >> k;
    long long a[n * k];
    for (int i = 0; i < n * k; i++) cin >> a[i];
    pair<long long, long long> ans[n + 1];
    map<long long, long long> last;
    map<long long, long long> occ;
    long long temp = 2;
    long long val = n / (k - 1);
    if (n % (k - 1) != 0) val++;
    long long curr = 0;
    for (int i = 0; i < n * k; i++) {
      if (last[a[i]] == -1) continue;
      occ[a[i]]++;
      if (occ[a[i]] == temp) {
        ans[a[i]] = make_pair(last[a[i]], i);
        last[a[i]] = -1;
        curr++;
        if (curr % val == 0) temp++;
      } else {
        last[a[i]] = i;
      }
    }
    for (int i = 1; i <= n; i++)
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << "\n";
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}

```