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 = input()
n = n.split()
m = int(n[1])
k = int(n[2])
n = int(n[0])
kol = m * n
s = str(kol - (k - 1) * 2) #kolvo v hvoste
i = 1; j = 1 # начальная точка таблицы
ki = 1 # направление прироста i (-1,1)
while k > 1:
    k -= 1
    print(2, i, j, end=' ')
    if (i == n and ki == 1) or (i == 1 and ki ==-1):
        ki *= -1
        j += 1
    else:
        i += ki
    print(i, j, end=' ')
    if (i == n and ki == 1) or (i == 1 and ki ==-1):
        ki *= -1
        j += 1
    else:
        i += ki
    print()

# остался хвост
while not(i == n and j == m):
    s = s + ' ' + str(i) + ' ' + str(j)
    if (i == n and ki == 1) or (i == 1 and ki ==-1):
        ki *= -1
        j += 1
    else:
        i += ki

print( s)