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>
import sys
input = sys.stdin.readline
from heapq import *

def main():
    n, k = map(int, input().split())
    clst = list(map(int, input().split()))
    max_ = (n + k - 1 - 1) // (k - 1)
    cnt = [0] * (n * k)
    aft = [-1] * n
    nex = [-1] * (n * k)
    bef = [-1] * (n * k)
    for i in range(n * k - 1, -1, -1):
        c = clst[i] - 1
        nex[i] = aft[c]
        aft[c] = i
    used = [False] * n
    hq = []
    for i in range(n * k):
        c = clst[i] - 1
        if not used[c]:
            used[c] = True
            j = nex[i]
            bef[j] = i
            heappush(hq, -j)
            for l in range(i, j + 1):
                cnt[l] += 1
        
        while cnt[i] > max_:
            j = -heappop(hq)
            used[clst[j] - 1] = False
            i = bef[j]
            for l in range(i, j + 1):
                cnt[l] -= 1
    ans = [-1] * n
    for h in hq:
        h = -h
        ans[clst[h] - 1] = (bef[h] + 1, h + 1)
    for row in ans:
        print(*row)
            
    
    
        
for _ in range(1):
    main()