#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void mine(T1 &x, const T2 &y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void maxe(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &b) {
  for (auto &k : b) os << k << ' ';
  return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &b) {
  for (auto &k : b) is >> k;
  return is;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 101;
long long last[N];
long long a[N * N];
pair<long long, long long> ans[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  memset(last, -1, sizeof(last));
  long long can = (n + (k - 2)) / (k - 1);
  for (long long i = 0; i < n; ++i) ans[i] = {-1, -1};
  for (long long i = 0; i < n * k; ++i) {
    long long x;
    cin >> x;
    --x;
    if (last[x] == -1) {
      last[x] = i;
    } else if (ans[x].first == -1) {
      long long mx = 0;
      for (long long j = last[x]; j <= i; ++j) {
        maxe(mx, a[j]);
      }
      if (mx < can) {
        for (long long j = last[x]; j <= i; ++j) {
          ++a[j];
        }
        ans[x] = {last[x] + 1, i + 1};
      }
      last[x] = i;
    }
  }
  for (long long i = 0; i < n; ++i) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}

### 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;
template <typename T1, typename T2>
inline void mine(T1 &x, const T2 &y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void maxe(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &b) {
  for (auto &k : b) os << k << ' ';
  return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &b) {
  for (auto &k : b) is >> k;
  return is;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long N = 101;
long long last[N];
long long a[N * N];
pair<long long, long long> ans[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  memset(last, -1, sizeof(last));
  long long can = (n + (k - 2)) / (k - 1);
  for (long long i = 0; i < n; ++i) ans[i] = {-1, -1};
  for (long long i = 0; i < n * k; ++i) {
    long long x;
    cin >> x;
    --x;
    if (last[x] == -1) {
      last[x] = i;
    } else if (ans[x].first == -1) {
      long long mx = 0;
      for (long long j = last[x]; j <= i; ++j) {
        maxe(mx, a[j]);
      }
      if (mx < can) {
        for (long long j = last[x]; j <= i; ++j) {
          ++a[j];
        }
        ans[x] = {last[x] + 1, i + 1};
      }
      last[x] = i;
    }
  }
  for (long long i = 0; i < n; ++i) {
    cout << ans[i].first << ' ' << ans[i].second << '\n';
  }
}

```