#include <bits/stdc++.h>
using namespace std;
namespace debug {
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
char elo(...);
template <class c>
auto elo(c* a) -> decltype(cerr << *a, 0);
struct stream {
  ~stream() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof elo<c>(0) != 1, stream&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof elo<c>(0) == 1, stream&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class a, class b>
  stream& operator<<(pair<a, b> p) {
    return *this << "(" << p.first << ", " << p.second << ")";
  }
  template <class c>
  stream& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; it++) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
  stream& _dbg(const string& s, int i, int b) { return *this; }
  template <class c, class... cs>
  stream& _dbg(const string& s, int i, int b, c arg, cs... args) {
    if (i == s.size()) return (*this << ": " << arg << "");
    b += (s[i] == '(') + (s[i] == '[') + (s[i] == '{');
    b -= (s[i] == ')') + (s[i] == ']') + (s[i] == '}');
    if (s[i] == ',' && b == 0)
      return (*this << ": " << arg << "     ")._dbg(s, i + 1, b, args...);
    return (s[i] == ' ' ? *this : *this << s[i])
        ._dbg(s, i + 1, b, arg, args...);
  }
};
}  // namespace debug
template <class p, class q>
pair<p, q> operator-(pair<p, q> a, pair<p, q> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <class p, class q>
pair<p, q> operator+(pair<p, q> a, pair<p, q> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <class p, class q>
void umin(p& a, const q& b) {
  if (a > b) a = b;
}
template <class p, class q>
void umax(p& a, const q& b) {
  if (a < b) a = b;
}
using lf = long double;
using ll = long long;
using cll = const ll;
using cint = const int;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int N = 1000005;
int t[N];
int last[N];
bool done[N];
pair<int, int> out[N];
void solve() {
  int n, i, j, k;
  scanf("%d%d", &n, &k);
  vector<pair<int, int> > vec;
  for (i = 1; i <= n * k; i++) {
    scanf("%d", &t[i]);
    if (last[t[i]] != 0) vec.push_back({i, last[t[i]]});
    last[t[i]] = i;
  }
  int lim = (n + k - 2) / (k - 1);
  sort(vec.begin(), vec.end());
  vector<pair<int, int> > res;
  for (pair<int, int> p : vec) {
    int a = p.second;
    int b = p.first;
    int col = t[a];
    assert(t[a] == t[b]);
    if (done[col] == true) continue;
    int cnt = 0;
    for (pair<int, int> p : res)
      if (p.second > a) cnt++;
    if (cnt >= lim) continue;
    res.push_back({a, b});
    out[col] = {a, b};
    done[col] = true;
  }
  assert(res.size() == n);
  for (i = 1; i <= n; i++) printf("%d %d\n", out[i].first, out[i].second);
}
int main() {
  int q = 1;
  while (q--) solve();
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
namespace debug {
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
char elo(...);
template <class c>
auto elo(c* a) -> decltype(cerr << *a, 0);
struct stream {
  ~stream() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof elo<c>(0) != 1, stream&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof elo<c>(0) == 1, stream&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class a, class b>
  stream& operator<<(pair<a, b> p) {
    return *this << "(" << p.first << ", " << p.second << ")";
  }
  template <class c>
  stream& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; it++) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
  stream& _dbg(const string& s, int i, int b) { return *this; }
  template <class c, class... cs>
  stream& _dbg(const string& s, int i, int b, c arg, cs... args) {
    if (i == s.size()) return (*this << ": " << arg << "");
    b += (s[i] == '(') + (s[i] == '[') + (s[i] == '{');
    b -= (s[i] == ')') + (s[i] == ']') + (s[i] == '}');
    if (s[i] == ',' && b == 0)
      return (*this << ": " << arg << "     ")._dbg(s, i + 1, b, args...);
    return (s[i] == ' ' ? *this : *this << s[i])
        ._dbg(s, i + 1, b, arg, args...);
  }
};
}  // namespace debug
template <class p, class q>
pair<p, q> operator-(pair<p, q> a, pair<p, q> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <class p, class q>
pair<p, q> operator+(pair<p, q> a, pair<p, q> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <class p, class q>
void umin(p& a, const q& b) {
  if (a > b) a = b;
}
template <class p, class q>
void umax(p& a, const q& b) {
  if (a < b) a = b;
}
using lf = long double;
using ll = long long;
using cll = const ll;
using cint = const int;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int N = 1000005;
int t[N];
int last[N];
bool done[N];
pair<int, int> out[N];
void solve() {
  int n, i, j, k;
  scanf("%d%d", &n, &k);
  vector<pair<int, int> > vec;
  for (i = 1; i <= n * k; i++) {
    scanf("%d", &t[i]);
    if (last[t[i]] != 0) vec.push_back({i, last[t[i]]});
    last[t[i]] = i;
  }
  int lim = (n + k - 2) / (k - 1);
  sort(vec.begin(), vec.end());
  vector<pair<int, int> > res;
  for (pair<int, int> p : vec) {
    int a = p.second;
    int b = p.first;
    int col = t[a];
    assert(t[a] == t[b]);
    if (done[col] == true) continue;
    int cnt = 0;
    for (pair<int, int> p : res)
      if (p.second > a) cnt++;
    if (cnt >= lim) continue;
    res.push_back({a, b});
    out[col] = {a, b};
    done[col] = true;
  }
  assert(res.size() == n);
  for (i = 1; i <= n; i++) printf("%d %d\n", out[i].first, out[i].second);
}
int main() {
  int q = 1;
  while (q--) solve();
  return 0;
}

```