1552_E. Colors and Intervals_70
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
import math

if sys.subversion[0] == "PyPy":
    import io, atexit

    sys.stdout = io.BytesIO()
    atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))

    sys.stdin = io.BytesIO(sys.stdin.read())
    input = lambda: sys.stdin.readline().rstrip()

#t = int(raw_input())

# python codesforces210710.py < input.txt
#for x in range(t):
str1 = raw_input()
str2 = raw_input()
n, k = str1.split(' ')
li = str2.split(' ')


n = int(n)
k = int(k)
li = map(int, li)
matt = []
#ind = 0
setdone = set()
rr = list(range(1,n+1))
biD = dict.fromkeys(rr)
bigfDD = biD.copy()

for i, l in enumerate(li):
    matt.append([i+1, l])
#matt2 = [x for x in matt if x[1] != 1]
#print(matt)
#print(matt2)


while len(matt) > 0:
    #print(matt)
    bigD = biD.copy()
    for i in range(0, len(matt)):
        #print(i)
        if matt[i][1] not in setdone:
            if bigD.get(matt[i][1]):
                bigfDD[matt[i][1]] = str(bigD.get(matt[i][1])) + ' ' + str(matt[i][0])
                v = matt[i][1]
                setdone.add(v)
                bigD = biD.copy()
            else:
                bigD[matt[i][1]] = matt[i][0]

                #print(bigD[matt[i][1]])
    matt = [x for x in matt if x[1] not in setdone]



for i in range(1,n+1):
    print(bigfDD[i])
#print(n, k)
#print(matt)

