#include <bits/stdc++.h>
using namespace std;
const int N = 100;
inline int read() {
  int ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
inline long long readll() {
  long long ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
struct node {
  int l, r, col;
} ans[N + 15], c[N * N + 15];
inline bool cmp(node e, node f) {
  if (e.r == f.r) return e.l < f.l;
  return e.r < f.r;
}
inline bool cmp1(node e, node f) { return e.col < f.col; }
int a[N * N + 15], pre[N * N + 15];
int vis[N + 15];
int cnt[N * N + 15];
int main() {
  int n = read();
  int k = read();
  int mxcnt = n / (k - 1) + (n % (k - 1) != 0);
  int tot = 0;
  for (register int i = 1; i <= n * k; i++) {
    a[i] = read();
    if (pre[a[i]]) {
      c[++tot] = (node){pre[a[i]], i, a[i]};
    }
    pre[a[i]] = i;
  }
  sort(c + 1, c + tot + 1, cmp);
  int sum = 0;
  for (register int i = 1; i <= tot; i++) {
    if (vis[c[i].col]) continue;
    bool flag = 1;
    for (register int j = c[i].l; j <= c[i].r; j++) {
      if (cnt[j] == mxcnt) {
        flag = 0;
        break;
      }
    }
    if (!flag) continue;
    ans[++sum] = c[i];
    vis[c[i].col] = 1;
    for (register int j = c[i].l; j <= c[i].r; j++) {
      cnt[j]++;
    }
  }
  sort(ans + 1, ans + sum + 1, cmp1);
  for (register int i = 1; i <= sum; i++) {
    printf("%d %d\n", ans[i].l, ans[i].r);
  }
  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;
const int N = 100;
inline int read() {
  int ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
inline long long readll() {
  long long ret = 0, op = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret * op;
}
void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
struct node {
  int l, r, col;
} ans[N + 15], c[N * N + 15];
inline bool cmp(node e, node f) {
  if (e.r == f.r) return e.l < f.l;
  return e.r < f.r;
}
inline bool cmp1(node e, node f) { return e.col < f.col; }
int a[N * N + 15], pre[N * N + 15];
int vis[N + 15];
int cnt[N * N + 15];
int main() {
  int n = read();
  int k = read();
  int mxcnt = n / (k - 1) + (n % (k - 1) != 0);
  int tot = 0;
  for (register int i = 1; i <= n * k; i++) {
    a[i] = read();
    if (pre[a[i]]) {
      c[++tot] = (node){pre[a[i]], i, a[i]};
    }
    pre[a[i]] = i;
  }
  sort(c + 1, c + tot + 1, cmp);
  int sum = 0;
  for (register int i = 1; i <= tot; i++) {
    if (vis[c[i].col]) continue;
    bool flag = 1;
    for (register int j = c[i].l; j <= c[i].r; j++) {
      if (cnt[j] == mxcnt) {
        flag = 0;
        break;
      }
    }
    if (!flag) continue;
    ans[++sum] = c[i];
    vis[c[i].col] = 1;
    for (register int j = c[i].l; j <= c[i].r; j++) {
      cnt[j]++;
    }
  }
  sort(ans + 1, ans + sum + 1, cmp1);
  for (register int i = 1; i <= sum; i++) {
    printf("%d %d\n", ans[i].l, ans[i].r);
  }
  return 0;
}

```