Problem:
Score : 300 points

Problem Statement

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is ., and black if c_{i,j} is #.

Consider doing the following operation:

Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.

You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Constraints

1 \leq H, W \leq 6

1 \leq K \leq HW

c_{i,j} is . or #.

Input

Input is given from Standard Input in the following format:

H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W}

Output

Print an integer representing the number of choices of rows and columns satisfying the condition.

Sample Input 1

2 3 2
..#
###

Sample Output 1

5

Five choices below satisfy the condition.

The 1-st row and 1-st column

The 1-st row and 2-nd column

The 1-st row and 3-rd column

The 1-st and 2-nd column

The 3-rd column

Sample Input 2

2 3 4
..#
###

Sample Output 2

1

One choice, which is choosing nothing, satisfies the condition.

Sample Input 3

2 2 3
##
##

Sample Output 3

0

Sample Input 4

6 6 8
..##..
.#..#.
#....#
######
#....#
#....#

Sample Output 4

208

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("Ofast")
#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define rep2(i,n) for(int i=0;i<=n;i++)
#define repr(i,a,n) for(int i=a;i<n;i++)
#define all(a) a.begin(),a.end()
#define P pair<long long,long long>
#define uni(e) e.erase(unique(e.begin(),e.end()),e.end())
#define double long double
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int INF=1e10;
int MOD=1e9+7;
template<class T> void out(T a){cout<<a<<'\n';}
template<class T> void yesno(T b){if(b)out("yes");else out("no");}
template<class T> void YesNo(T b){if(b)out("Yes");else out("No");}
template<class T> void YESNO(T b){if(b)out("YES");else out("NO");}
template<class T> void noyes(T b){if(b)out("no");else out("yes");}
template<class T> void NoYes(T b){if(b)out("No");else out("Yes");}
template<class T> void NOYES(T b){if(b)out("NO");else out("YES");}
int keta(int a){
  double b=a;
  b=log10(b);
  int c=b;
  return c+1;
}
int kurai(int a,int b){
  int m=pow(10,b),n=pow(10,b-1);
  return (a%m)/n;
}
int ketawa(int a){
  int k=0;
  rep(i,keta(a)){
  k+=kurai(a,i+1);
  }
  return k;
}
int gcd(int a,int b){
  if(a%b==0)
    return b;
  return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
int mid(int a,int b,int c){
  vector<int> k(3);
  k[0]=a;k[1]=b;k[2]=c;
  sort(all(k));
  return k[1];
}
bool sosuu(int a){
  bool b=1;
  if(a<=1)
    return 0;
  else{
  rep1(i,sqrt(a)-1){
  if(a%(i+1)==0)
  b=0;
  }
  return b;
  }
}
int modpow(int a,int b,int m=MOD){
 if(!b)return 1; if(b&1)return modpow(a,b-1,m)*a%m;
 int memo = modpow(a,b>>1,m);
 return memo*memo%m;
}
int fact_mod(int n) {
    int f=1; 
for(int i=2;i<n+1;i++) f=f*(i%MOD)% MOD;
    return f;
}
int mod_pow(int x,int n) {
    int res=1;
    while(n>0){
        if(n&1) res=(res*x)%MOD; 
        x=(x*x)%MOD;        
n>>=1; 
    }
    return res;
}
int c_mod(int n, int r) {
    if(r>n-r) r=n-r;
    if(r==0) return 1;
    int a=1;
    rep(i,r) 
a=a*((n-i)%MOD)%MOD;
    int b=mod_pow(fact_mod(r), MOD-2);
    return (a%MOD)*(b%MOD)%MOD;
}
signed main(){
  int a,b,c;
  cin>>a>>b>>c;
  char d[a][b];
  rep(i,a)rep(j,b)
    cin>>d[i][j];
  int h=0;
  int p[a][b];
  rep(i,a)rep(s,b){
  p[i][s]=d[i][s];
  }
  rep(i,(1<<6)){
  bitset<6> e(i);
  rep(j,(1<<6)){
  bitset<6> f(j);
    rep(k,a){
      if(e.test(k)){
      rep(l,b){
      d[k][l]='.';
      }
      }
    }
    rep(m,b){
      if(f.test(m)){
      rep(n,a){
      d[n][m]='.';
      }
      }
    }
    int t=0;
    rep(k,a)rep(m,b){
    if(d[k][m]=='#')
      t++;
    }
    if(t==c)
      h++;
     rep(k,a)rep(l,b)
    d[k][l]=p[k][l];
  }
  }
  out(h);
}