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.
"""
Codeforces Round 252 Div 2 Problem C

Author  : chaotic_iak
Language: Python 3.3.4
"""

def read(mode=2):
    # 0: String
    # 1: List of strings
    # 2: List of integers
    inputs = input().strip()
    if mode == 0:
        return inputs
    if mode == 1:
        return inputs.split()
    if mode == 2:
        return [int(x) for x in inputs.split()]

def write(s="\n"):
    if isinstance(s, list): s = " ".join(s)
    s = str(s)
    print(s, end="")

################################################### SOLUTION
n,m,k = read()

cells = []
for i in range(n):
    if not i%2:
        for j in range(m): cells.append((i+1,j+1))
    else:
        for j in range(m-1, -1, -1): cells.append((i+1,j+1))

for i in range(k-1):
    a = cells.pop(0)
    b = cells.pop(0)
    print(2, a[0], a[1], b[0], b[1])

print(len(cells), end=" ")
for i in cells:
    print(i[0], i[1], end=" ")