#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int* array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int* array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int>& vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int>& vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v1) {
  int n = v1.size();
  for (int i = 0; i < n; i++) {
    if (i) os << " ";
    os << v1[i];
  }
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v1) {
  int n = v1.size();
  for (int i = 0; i < n; i++) is >> v1[i];
  return is;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
  is >> p.first >> p.second;
  return is;
}
namespace sol {
void solve() {
  int n, m;
  int i, j, k;
  int a, b, c;
  cin >> n >> m;
  vector<int> v1, cnt(n), bef(n);
  vector<char> used(n);
  vector<pair<int, int> > vs(n);
  for (i = 0; i < n * m; i++) {
    cin >> a;
    v1.push_back(a - 1);
  }
  for (i = 0; i < (n - 1) / (m - 1) + 1; i++) {
    int l = i * (m - 1), r = min((i + 1) * (m - 1), n);
    a = 2;
    for (j = 0; j < n * m; j++) {
      if (v1[j] < l || r <= v1[j]) continue;
      cnt[v1[j]]++;
      if (used[v1[j]]) continue;
      if (cnt[v1[j]] != a) {
        bef[v1[j]] = j;
        continue;
      }
      used[v1[j]] = 1;
      vs[v1[j]] = {bef[v1[j]] + 1, j + 1};
      a++;
    }
  }
  for (i = 0; i < n; i++) {
    cout << vs[i] << endl;
  }
}
}  // namespace sol
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  sol::solve();
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int* array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int* array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int>& vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int>& vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v1) {
  int n = v1.size();
  for (int i = 0; i < n; i++) {
    if (i) os << " ";
    os << v1[i];
  }
  return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v1) {
  int n = v1.size();
  for (int i = 0; i < n; i++) is >> v1[i];
  return is;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
  is >> p.first >> p.second;
  return is;
}
namespace sol {
void solve() {
  int n, m;
  int i, j, k;
  int a, b, c;
  cin >> n >> m;
  vector<int> v1, cnt(n), bef(n);
  vector<char> used(n);
  vector<pair<int, int> > vs(n);
  for (i = 0; i < n * m; i++) {
    cin >> a;
    v1.push_back(a - 1);
  }
  for (i = 0; i < (n - 1) / (m - 1) + 1; i++) {
    int l = i * (m - 1), r = min((i + 1) * (m - 1), n);
    a = 2;
    for (j = 0; j < n * m; j++) {
      if (v1[j] < l || r <= v1[j]) continue;
      cnt[v1[j]]++;
      if (used[v1[j]]) continue;
      if (cnt[v1[j]] != a) {
        bef[v1[j]] = j;
        continue;
      }
      used[v1[j]] = 1;
      vs[v1[j]] = {bef[v1[j]] + 1, j + 1};
      a++;
    }
  }
  for (i = 0; i < n; i++) {
    cout << vs[i] << endl;
  }
}
}  // namespace sol
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  sol::solve();
}

```