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.
def fucn(x,y,c):
    if x%2:
        y= y+1
        if y>c:
            x=x+1
            y=c
    else:
        y= y-1
        if y<1:
            x=x+1;
            y=1;
    return x,y

scann=input()
x=1
y=1
acum=0
arr=[]
numbers=scann.split()
row= numbers[0]
column= numbers[1]
tube = numbers [2]
r= int(row)
c= int(column)
t= int(tube)
table= r*c
aux= (int(table))/t
for i in range(t-1):
    arr.append(int(aux))
    for j in range(int(aux)):
        arr.append(int(x))
        arr.append(int(y))
        x,y= fucn(x,y,c)
        acum=acum+1
    print(*arr)
    arr=[]
arr.append(table-acum)
new= table-acum
for k in range(int(new)):
    arr.append(int(x))
    arr.append(int(y))
    x,y= fucn(x,y,c)
print(*arr)