#include <bits/stdc++.h>
using namespace std;
template <typename T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = (1ll << 31) - 1;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int T = 1;
  for (int(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n, k;
    cin >> n >> k;
    vector<long long> pos[n + 1];
    for (int(p) = (0); (p) <= ((n * k) - 1); (p)++) {
      long long col;
      cin >> col;
      pos[col].emplace_back(p + 1);
    }
    vector<tuple<long long, long long, long long>> seg;
    for (int(col) = (1); (col) <= (n); (col)++) {
      for (int(i) = (0); (i) <= ((k - 1) - 1); (i)++) {
        seg.emplace_back(pos[col][i + 1], pos[col][i], col);
      }
    }
    sort((seg).begin(), (seg).end());
    long long tot = 0;
    vector<long long> done(n + 1, 0);
    vector<pair<long long, long long>> ans(n + 1);
    while (tot < n) {
      long long idx = 0, crt = 0;
      while (idx < seg.size()) {
        auto [r, l, col] = seg[idx++];
        if (l <= crt || done[col]) continue;
        done[col] = 1, tot++;
        ans[col] = {l, r};
        crt = r;
      }
    }
    for (int(col) = (1); (col) <= (n); (col)++) {
      auto [l, r] = ans[col];
      ;
      cout << l << " " << r << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 T>
using MinHeap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using MaxHeap = priority_queue<T>;
const long double PI = 3.14159265358979323846;
const long double E = 2.718281828459;
long long MOD = (1ll << 31) - 1;
long long M(long long n) { return (n % MOD + MOD) % MOD; }
long long pow2(long long n) { return 1ll << n; }
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int main(int argn, char **argv) {
  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int T = 1;
  for (int(I) = (0); (I) <= ((T)-1); (I)++) {
    long long n, k;
    cin >> n >> k;
    vector<long long> pos[n + 1];
    for (int(p) = (0); (p) <= ((n * k) - 1); (p)++) {
      long long col;
      cin >> col;
      pos[col].emplace_back(p + 1);
    }
    vector<tuple<long long, long long, long long>> seg;
    for (int(col) = (1); (col) <= (n); (col)++) {
      for (int(i) = (0); (i) <= ((k - 1) - 1); (i)++) {
        seg.emplace_back(pos[col][i + 1], pos[col][i], col);
      }
    }
    sort((seg).begin(), (seg).end());
    long long tot = 0;
    vector<long long> done(n + 1, 0);
    vector<pair<long long, long long>> ans(n + 1);
    while (tot < n) {
      long long idx = 0, crt = 0;
      while (idx < seg.size()) {
        auto [r, l, col] = seg[idx++];
        if (l <= crt || done[col]) continue;
        done[col] = 1, tot++;
        ans[col] = {l, r};
        crt = r;
      }
    }
    for (int(col) = (1); (col) <= (n); (col)++) {
      auto [l, r] = ans[col];
      ;
      cout << l << " " << r << endl;
    }
  }
  return 0;
}

```