#include <bits/stdc++.h>
using namespace std;
vector<int> vec;
int n, k;
const int dydis = 101;
int nxt[dydis * dydis + 10];
vector<pair<int, int>> transform(vector<int> &mas, vector<int> &take) {
  vector<pair<int, int>> ret;
  int i = 0;
  for (auto &x : mas) {
    bool tk = 0;
    for (auto &y : take) {
      if (y == x) tk = 1;
    }
    if (tk) ret.push_back({x, i});
    i++;
  }
  return ret;
}
vector<pair<int, int>> transform(vector<pair<int, int>> &mas, int ign, int st) {
  vector<pair<int, int>> ret;
  pair<int, int> x;
  for (int i = st; i < mas.size(); i++) {
    x = mas[i];
    if (x.first != ign) {
      ret.push_back(x);
    }
  }
  return ret;
}
vector<pair<int, int>> fAns(int l, int r) {
  vector<int> st;
  for (int i = l; i <= r; i++) {
    st.push_back(i);
  }
  auto mas = transform(vec, st);
  vector<pair<int, int>> ret;
  for (int i = 0; i < st.size(); i++) {
    for (auto x : mas) {
    }
    pair<int, int> mn = {dydis * dydis, dydis * dydis};
    for (int j = 0; j < mas.size(); j++) {
      mn = min(mn, make_pair(nxt[mas[j].second], mas[j].second));
    }
    int tol = 0;
    for (int j = 0; j < mas.size(); j++) {
      if (mn.first == mas[j].second) tol = j;
    }
    ret.push_back({mn.second, mn.first});
    mas = transform(mas, vec[mn.first], tol + 1);
  }
  return ret;
}
int main() {
  srand(time(0));
  cin >> n >> k;
  int grupiu = (n / (k - 1)) + (n % (k - 1) != 0);
  int grupeje = (n) / grupiu;
  vector<int> mas(grupiu, grupeje);
  int sm = grupeje * grupiu;
  for (int i = 0; i < grupiu; i++) {
    if (sm != n) {
      mas[i]++;
      sm++;
    }
  }
  for (int i = 0; i < n * k; i++) {
    int a;
    cin >> a;
    vec.push_back(a - 1);
  }
  for (int i = 0; i < vec.size(); i++) {
    int nx = n * k;
    for (int j = i + 1; j < vec.size(); j++) {
      if (vec[j] == vec[i]) {
        nx = j;
        break;
      }
    }
    nxt[i] = nx;
  }
  int l = 0;
  vector<pair<int, pair<int, int>>> ans;
  for (int i = 0; i < mas.size(); i++) {
    auto st = fAns(l, l + mas[i] - 1);
    l += mas[i];
    for (auto x : st) ans.push_back({vec[x.first], x});
  }
  sort(ans.begin(), ans.end());
  for (auto x : ans) {
    cout << x.second.first + 1 << " " << x.second.second + 1 << "\n";
    if (vec[x.second.first] != vec[x.second.second]) {
      cout << "NO";
      return 0;
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
vector<int> vec;
int n, k;
const int dydis = 101;
int nxt[dydis * dydis + 10];
vector<pair<int, int>> transform(vector<int> &mas, vector<int> &take) {
  vector<pair<int, int>> ret;
  int i = 0;
  for (auto &x : mas) {
    bool tk = 0;
    for (auto &y : take) {
      if (y == x) tk = 1;
    }
    if (tk) ret.push_back({x, i});
    i++;
  }
  return ret;
}
vector<pair<int, int>> transform(vector<pair<int, int>> &mas, int ign, int st) {
  vector<pair<int, int>> ret;
  pair<int, int> x;
  for (int i = st; i < mas.size(); i++) {
    x = mas[i];
    if (x.first != ign) {
      ret.push_back(x);
    }
  }
  return ret;
}
vector<pair<int, int>> fAns(int l, int r) {
  vector<int> st;
  for (int i = l; i <= r; i++) {
    st.push_back(i);
  }
  auto mas = transform(vec, st);
  vector<pair<int, int>> ret;
  for (int i = 0; i < st.size(); i++) {
    for (auto x : mas) {
    }
    pair<int, int> mn = {dydis * dydis, dydis * dydis};
    for (int j = 0; j < mas.size(); j++) {
      mn = min(mn, make_pair(nxt[mas[j].second], mas[j].second));
    }
    int tol = 0;
    for (int j = 0; j < mas.size(); j++) {
      if (mn.first == mas[j].second) tol = j;
    }
    ret.push_back({mn.second, mn.first});
    mas = transform(mas, vec[mn.first], tol + 1);
  }
  return ret;
}
int main() {
  srand(time(0));
  cin >> n >> k;
  int grupiu = (n / (k - 1)) + (n % (k - 1) != 0);
  int grupeje = (n) / grupiu;
  vector<int> mas(grupiu, grupeje);
  int sm = grupeje * grupiu;
  for (int i = 0; i < grupiu; i++) {
    if (sm != n) {
      mas[i]++;
      sm++;
    }
  }
  for (int i = 0; i < n * k; i++) {
    int a;
    cin >> a;
    vec.push_back(a - 1);
  }
  for (int i = 0; i < vec.size(); i++) {
    int nx = n * k;
    for (int j = i + 1; j < vec.size(); j++) {
      if (vec[j] == vec[i]) {
        nx = j;
        break;
      }
    }
    nxt[i] = nx;
  }
  int l = 0;
  vector<pair<int, pair<int, int>>> ans;
  for (int i = 0; i < mas.size(); i++) {
    auto st = fAns(l, l + mas[i] - 1);
    l += mas[i];
    for (auto x : st) ans.push_back({vec[x.first], x});
  }
  sort(ans.begin(), ans.end());
  for (auto x : ans) {
    cout << x.second.first + 1 << " " << x.second.second + 1 << "\n";
    if (vec[x.second.first] != vec[x.second.second]) {
      cout << "NO";
      return 0;
    }
  }
  return 0;
}

```