#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
void chkmax(T &first, T second) {
  first < second ? first = second : T();
}
template <typename T>
void chkmin(T &first, T second) {
  second < first ? first = second : T();
}
template <typename T>
void readint(T &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
  first *= f;
}
template <typename T>
T divceil(T first, T second) {
  return (first - 1) / second + 1;
}
int n, k, c[10005], last[105], cnt[105], f[105];
struct interval {
  interval(int _c, int _l, int _r) : c(_c), l(_l), r(_r) {}
  interval() = default;
  int c, l, r;
};
vector<interval> V[105];
interval ans[105];
bool cmp(interval &i1, interval &i2) { return i1.r < i2.r; }
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    last[i] = -1;
    cnt[i] = 0, f[i] = 0;
  }
  for (int i = 0; i < n * k; ++i) {
    readint(c[i]), --c[i];
    if (last[c[i]] >= 0) {
      V[cnt[c[i]]++].push_back(interval(c[i], last[c[i]] + 1, i + 1));
    }
    last[c[i]] = i;
  }
  memset(f, 0, sizeof(f));
  int b = divceil(n, k - 1);
  for (int i = 0; i < k; ++i) {
    sort(V[i].begin(), V[i].end(), cmp);
    int n_selected = 0;
    for (auto &itv : V[i]) {
      if (!f[itv.c] && n_selected < b) {
        ans[itv.c] = itv;
        f[itv.c] = 1;
        ++n_selected;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d %d\n", ans[i].l, ans[i].r);
  }
  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("-Ofast")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <typename T>
void chkmax(T &first, T second) {
  first < second ? first = second : T();
}
template <typename T>
void chkmin(T &first, T second) {
  second < first ? first = second : T();
}
template <typename T>
void readint(T &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
  first *= f;
}
template <typename T>
T divceil(T first, T second) {
  return (first - 1) / second + 1;
}
int n, k, c[10005], last[105], cnt[105], f[105];
struct interval {
  interval(int _c, int _l, int _r) : c(_c), l(_l), r(_r) {}
  interval() = default;
  int c, l, r;
};
vector<interval> V[105];
interval ans[105];
bool cmp(interval &i1, interval &i2) { return i1.r < i2.r; }
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    last[i] = -1;
    cnt[i] = 0, f[i] = 0;
  }
  for (int i = 0; i < n * k; ++i) {
    readint(c[i]), --c[i];
    if (last[c[i]] >= 0) {
      V[cnt[c[i]]++].push_back(interval(c[i], last[c[i]] + 1, i + 1));
    }
    last[c[i]] = i;
  }
  memset(f, 0, sizeof(f));
  int b = divceil(n, k - 1);
  for (int i = 0; i < k; ++i) {
    sort(V[i].begin(), V[i].end(), cmp);
    int n_selected = 0;
    for (auto &itv : V[i]) {
      if (!f[itv.c] && n_selected < b) {
        ans[itv.c] = itv;
        f[itv.c] = 1;
        ++n_selected;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d %d\n", ans[i].l, ans[i].r);
  }
  return 0;
}

```