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 = [int(k) for k in input().split()]
total = n * m
length = [int(total/k)]*(k-1)
length.append(total - (int(total/k))*(k-1))
#print(length)
table =[[0]*(m+1) for _ in range(n+1)]
i = 1


path = []
for r in range(1, n+1):
    if r % 2 == 0: #odd rows, go forward
        for c in range(m, 0,-1):
            path.append(c)
            path.append(r)
    else: #even rows, go backwards
        for c in range(1, m+1):
            path.append(c)
            path.append(r)

#print(path)
i = 0    
for l in length:
    print(l, end = " ")
    s = ""
    for x in path[i:i+2*l]:
        s += str(x) + " "
    print(s)
    #print(path[i:i+2*l], end = " ")
    i = i+2*l