#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937_64 rnd(time(0));
long long MOD = 998244353;
const long double E = 1e-10;
long long n, m, k, T, nw[100001], a[100001];
pair<long long, long long> ans[100001];
long long binpow(long long h, long long r, long long md) {
  long long l = 1;
  while (r) {
    if (r & 1) l = (l * h) % md;
    h = (h * h) % md;
    r /= 2;
  }
  return l;
}
long long read() {
  char c = getchar();
  while (!isdigit(c)) {
    c = getchar();
  }
  long long res = 0;
  while (isdigit(c)) {
    res = res * 10 + c - '0';
    c = getchar();
  }
  return res;
}
long long gcd(long long x, long long y) {
  while (x && y) {
    if (x < y) swap(x, y);
    x %= y;
  }
  return x + y;
}
bool tt[1001];
vector<long long> c[1001];
vector<long long> v[1001];
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) c[i].clear(), tt[i] = 0;
  for (int i = 1; i <= n * k; i++) {
    cin >> a[i];
    c[a[i]].push_back(i);
  }
  vector<long long> z;
  for (int i = 1; i <= n; i++) {
    z.push_back(i);
  }
  while (1) {
    long long y = 0;
    for (int i = 1; i <= n; i++)
      if (!tt[i]) nw[i] = 0, y = 1;
    if (!y) break;
    long long p = 0;
    while (1) {
      long long mn = 1e18, pp = 0, u = 0;
      for (int i = 1; i <= n; i++)
        if (!tt[i]) {
          while (nw[i] < c[i].size() - 1 && c[i][nw[i]] <= p) nw[i]++;
          if (nw[i] < c[i].size() - 1 && c[i][nw[i] + 1] < mn)
            mn = c[i][nw[i] + 1], u = c[i][nw[i]], pp = i;
        }
      if (pp == 0) break;
      tt[pp] = 1;
      ans[pp] = make_pair(u, mn);
      p = mn;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
}
int main() {
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize("-O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937_64 rnd(time(0));
long long MOD = 998244353;
const long double E = 1e-10;
long long n, m, k, T, nw[100001], a[100001];
pair<long long, long long> ans[100001];
long long binpow(long long h, long long r, long long md) {
  long long l = 1;
  while (r) {
    if (r & 1) l = (l * h) % md;
    h = (h * h) % md;
    r /= 2;
  }
  return l;
}
long long read() {
  char c = getchar();
  while (!isdigit(c)) {
    c = getchar();
  }
  long long res = 0;
  while (isdigit(c)) {
    res = res * 10 + c - '0';
    c = getchar();
  }
  return res;
}
long long gcd(long long x, long long y) {
  while (x && y) {
    if (x < y) swap(x, y);
    x %= y;
  }
  return x + y;
}
bool tt[1001];
vector<long long> c[1001];
vector<long long> v[1001];
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) c[i].clear(), tt[i] = 0;
  for (int i = 1; i <= n * k; i++) {
    cin >> a[i];
    c[a[i]].push_back(i);
  }
  vector<long long> z;
  for (int i = 1; i <= n; i++) {
    z.push_back(i);
  }
  while (1) {
    long long y = 0;
    for (int i = 1; i <= n; i++)
      if (!tt[i]) nw[i] = 0, y = 1;
    if (!y) break;
    long long p = 0;
    while (1) {
      long long mn = 1e18, pp = 0, u = 0;
      for (int i = 1; i <= n; i++)
        if (!tt[i]) {
          while (nw[i] < c[i].size() - 1 && c[i][nw[i]] <= p) nw[i]++;
          if (nw[i] < c[i].size() - 1 && c[i][nw[i] + 1] < mn)
            mn = c[i][nw[i] + 1], u = c[i][nw[i]], pp = i;
        }
      if (pp == 0) break;
      tt[pp] = 1;
      ans[pp] = make_pair(u, mn);
      p = mn;
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
}
int main() {
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```