#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<long long, long long>;
const double EPS = 1e-9;
const long long INF = 1e18;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
void debug_out(string s, T t) {
  cout << "[" << s << ": " << t << "]\n";
}
template <typename T, typename... U>
void debug_out(string s, T t, U... u) {
  long long w = 0, c = 0;
  while (w < (long long)s.size()) {
    if (s[w] == '(') c++;
    if (s[w] == ')') c--;
    if (!c and s[w] == ',') break;
    w++;
  }
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  debug_out(s.substr(w + 2, (long long)s.size() - w - 1), u...);
}
const long long N = 1e6 + 5;
vector<long long> idx[105];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> c(n * k + 1), prev(n * k + 1), done(n + 1);
  vector<pii> ans(n + 1);
  for (long long i = 1; i <= n * k; i++) cin >> c[i], idx[c[i]].push_back(i);
  bool flg = 1;
  while (true) {
    bool found = 0;
    for (long long &x : prev) x = 0;
    for (long long i = 1; i <= n * k; i++) {
      if (!done[c[i]]) {
        if (prev[c[i]] > 0) {
          done[c[i]] = 1, ans[c[i]].first = prev[c[i]], ans[c[i]].second = i,
          found = 1;
          for (long long &x : prev) x = 0;
        } else
          prev[c[i]] = i;
      }
    }
    if (!found) break;
  }
  for (long long i = 1; i <= n; i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  return 0;
}

### 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;
using ll = long long;
using pii = pair<long long, long long>;
const double EPS = 1e-9;
const long long INF = 1e18;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
void debug_out(string s, T t) {
  cout << "[" << s << ": " << t << "]\n";
}
template <typename T, typename... U>
void debug_out(string s, T t, U... u) {
  long long w = 0, c = 0;
  while (w < (long long)s.size()) {
    if (s[w] == '(') c++;
    if (s[w] == ')') c--;
    if (!c and s[w] == ',') break;
    w++;
  }
  cout << "[" << s.substr(0, w) << ": " << t << "] ";
  debug_out(s.substr(w + 2, (long long)s.size() - w - 1), u...);
}
const long long N = 1e6 + 5;
vector<long long> idx[105];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> c(n * k + 1), prev(n * k + 1), done(n + 1);
  vector<pii> ans(n + 1);
  for (long long i = 1; i <= n * k; i++) cin >> c[i], idx[c[i]].push_back(i);
  bool flg = 1;
  while (true) {
    bool found = 0;
    for (long long &x : prev) x = 0;
    for (long long i = 1; i <= n * k; i++) {
      if (!done[c[i]]) {
        if (prev[c[i]] > 0) {
          done[c[i]] = 1, ans[c[i]].first = prev[c[i]], ans[c[i]].second = i,
          found = 1;
          for (long long &x : prev) x = 0;
        } else
          prev[c[i]] = i;
      }
    }
    if (!found) break;
  }
  for (long long i = 1; i <= n; i++)
    cout << ans[i].first << " " << ans[i].second << "\n";
  return 0;
}

```