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.
from collections import defaultdict
n,m,k=map(int,input().split())
g=[]
for i in range(n):
    g.append([0]*m)
d=defaultdict(list)
c=1
i=0
j=0
inc=1
while i<n:
    if inc==1:
        if j+1<m:
            g[i][j]=c
            g[i][j+1]=c
            d[c].append(i+1)
            d[c].append(j+1)
            d[c].append(i+1)
            d[c].append(j+2)
            c=min(c+1,k)
            j+=2
        elif j==m-1:
            g[i][j]=c
            d[c].append(i+1)
            d[c].append(j+1)
            if i+1<n:
                g[i+1][j]=c
                d[c].append(i+2)
                d[c].append(j+1)
                c=min(c+1,k)
                j=m-2
            i=i+1
            inc=-1
        else:
            j=m-1
            i=i+1
            inc=-1
    else:
        if j-1>=0:
            g[i][j]=c
            g[i][j-1]=c
            d[c].append(i+1)
            d[c].append(j+1)
            d[c].append(i+1)
            d[c].append(j)
            c=min(c+1,k)
            j-=2
        elif j==0:
            g[i][j]=c
            d[c].append(i+1)
            d[c].append(j+1)
            if i+1<n:
                g[i+1][j]=c
                d[c].append(i+2)
                d[c].append(j+1)
                c=min(c+1,k)
                j=1
            i=i+1
            inc=1
        else:
            j=0
            inc=1
            i+=1
    if i==n:
        break
for i in range(1,k+1):
    print(len(d[i])//2,*d[i])