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,os,io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

n,k = [int(i) for i in input().split()]
a = [int(i)-1 for i in input().split()]

vis = [0]*(n)
cnt = [0]*(n*k)
ind = [[] for i in range (n)]
for i in range (len(a)):
    ind[a[i]].append(i)
b = []
for i in range (n):
    for j in range (1,len(ind[i])):
        b.append([ind[i][j-1], ind[i][j]])
b.sort(key = lambda x: x[1])
ans = []
maxi = (n + k-2)//(k-1)
for i in b:
    if vis[a[i[0]]]:
        continue
    mw = 0
    for j in range (i[0], i[1]+1):
        mw = max(mw, cnt[j])
    if mw < maxi:
        vis[a[i[0]]]=1
        for j in range (i[0], i[1]+1):
            cnt[j]+=1
        ans.append([a[i[0]], i[0]+1, i[1]+1])
ans = ans[:n]
ans.sort()
for i in ans:
    print(i[1], i[2])