Valera has got a rectangle table consisting of *n* rows and *m* columns. Valera numbered the table rows starting from one, from top to bottom and the columns – starting from one, from left to right. We will represent cell that is on the intersection of row *x* and column *y* by a pair of integers (*x*, *y*).

Valera wants to place exactly *k* tubes on his rectangle table. A tube is such sequence of table cells (*x*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">1</sub>), (*x*<sub class="lower-index">2</sub>, *y*<sub class="lower-index">2</sub>), ..., (*x*<sub class="lower-index">*r*</sub>, *y*<sub class="lower-index">*r*</sub>), that: 
 -  *r* ≥ 2; -  for any integer *i* (1 ≤ *i* ≤ *r* - 1) the following equation |*x*<sub class="lower-index">*i*</sub> - *x*<sub class="lower-index">*i* + 1</sub>| + |*y*<sub class="lower-index">*i*</sub> - *y*<sub class="lower-index">*i* + 1</sub>| = 1 holds; -  each table cell, which belongs to the tube, must occur exactly once in the sequence. 
Valera thinks that the tubes are arranged in a fancy manner if the following conditions are fulfilled: 
 -  no pair of tubes has common cells; -  each cell of the table belongs to some tube. 
Help Valera to arrange *k* tubes on his rectangle table in a fancy manner.
The first line contains three space-separated integers *n*, *m*, *k* (2 ≤ *n*, *m* ≤ 300; 2 ≤ 2*k* ≤ *n*·*m*) — the number of rows, the number of columns and the number of tubes, correspondingly.
n,m,k = list(map(int,input().split()))
unit_tube = (n * m) // k
long_tube = (n * m) % k + unit_tube

print(long_tube,end='')
i = 0
j = 0
ref = 0
for _ in range(long_tube):
    print(' {0} {1}'.format(i + 1,abs(j + 1 - ref)),end='')
    if((j + 1) % (m) == 0):
        i+=1
        ref = 0 if(ref) else m + 1
    j = (j + 1) % m
print()
for _ in range(k - 1):
    print(unit_tube,end='')
    for _ in range(unit_tube):
        print(' {0} {1}'.format(i + 1,abs(j + 1 - ref)),end='')
        if((j + 1) % (m) == 0):
            i+=1
            ref = 0 if(ref) else m + 1
        #j = abs((j + 1) % (m) - ref)
        j = (j + 1) % m
    print()
        