#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long cmb(long long r, long long c) {
  return (c > r || c < 0) ? 0 : fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
int vec[10000];
int ok[100];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n * k; i++) {
    scanf("%d", &vec[i]);
    vec[i]--;
  }
  vector<pair<int, int> > ans;
  ans.resize(n);
  map<int, int> ma;
  while (1) {
    int ff = 0;
    for (int i = 0; i < n; i++)
      if (ok[i] == 0) ff = 1;
    if (ff == 0) break;
    ma.clear();
    for (int i = 0; i < n * k; i++) {
      if (ok[vec[i]])
        continue;
      else {
        if (ma.find(vec[i]) == ma.end())
          ma[vec[i]] = i;
        else {
          ans[vec[i]] = {ma[vec[i]], i};
          ok[vec[i]] = 1;
          ma.clear();
        }
      }
    }
  }
  for (pair<int, int> i : ans) printf("%d %d\n", i.first + 1, i.second + 1);
}

### Prompt

Generate 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>
using namespace std;
const int MOD = 1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long cmb(long long r, long long c) {
  return (c > r || c < 0) ? 0 : fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
int vec[10000];
int ok[100];
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n * k; i++) {
    scanf("%d", &vec[i]);
    vec[i]--;
  }
  vector<pair<int, int> > ans;
  ans.resize(n);
  map<int, int> ma;
  while (1) {
    int ff = 0;
    for (int i = 0; i < n; i++)
      if (ok[i] == 0) ff = 1;
    if (ff == 0) break;
    ma.clear();
    for (int i = 0; i < n * k; i++) {
      if (ok[vec[i]])
        continue;
      else {
        if (ma.find(vec[i]) == ma.end())
          ma[vec[i]] = i;
        else {
          ans[vec[i]] = {ma[vec[i]], i};
          ok[vec[i]] = 1;
          ma.clear();
        }
      }
    }
  }
  for (pair<int, int> i : ans) printf("%d %d\n", i.first + 1, i.second + 1);
}

```