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: Accepted)

#include<bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define int long long 
#define pb push_back
#define ins insert
#define rs resize
#define br break;
#define cont continue;
#define vi vector<int>
#define vll vector<long long>
#define vvi vector<vector<int>>
#define si set<ll>
#define sll set<long long>
#define pii pair<ll,ll>
#define pll pair<long long, long long>
#define speed ios_base::sync_with_stdio(false);cin.tie(NULL);
#define vec vector
#define cend cout<<endl;
#define sc second
#define fr first
#define retz return 0;
#define ll_MAX (ll)(1e9*1e9)
#define ll_MIN (ll)(-1e9*1e9)
#define lb lower_bound
#define ub upper_bound
#define rep(i,a,b)  for(int i=a; i<=b; i++)
#define repd(i,a,b) for(int i=a; i>=b; i--)
#define ps(arr, n) for(int i=1; i<n; i++)arr[i]+=arr[i-1];
#define inp(arr, n)for(int i=0; i<n; i++)cin>>arr[i];
#define out(arr, n)for(int i=0; i<n; i++)cout<<arr[i];
#define inp1(arr,n,m)for(int i=0; i<n; i++)for(int j=0;j<m;j++)cin>>arr[i][j];
#define out1(arr,n,m)for(int i=0; i<n; i++)for(int j=0;j<m;j++)cin>>arr[i][j];
#define loop(q) for(auto it:q)
#define loop1(q) for(auto&it:q)
#define db1 cout << "*" << endl
#define db2 cout << "YES" << endl
using namespace std;
using namespace __gnu_pbds; 
#define oset(X) tree<X,null_type,less<X>, rb_tree_tag, tree_order_statistics_node_update> 
#define roset(X) tree<X,null_type,greater<X>, rb_tree_tag, tree_order_statistics_node_update>  
#define okey order_of_key //lower_bound_set
#define ofind find_by_order
#define cyes cout << "Yes" << endl; 
#define cno cout << "No" << endl;
#define mod (long long)1000000007
#define mod2 (long long)998244353
#define INT_MAX1 900000000
#define MAXN 2000000
#define INF INT_MAX



 
void solve()
{
    int n,m,k1;
    cin >> n >> m >> k1;
    string s[n];
    int ans = 0;
    inp(s, n);
    rep(i,0,(1<<n)-1)
        rep(j,0,(1<<m)-1)
        {
            string x[n];
            rep(i,0,n-1)
                x[i] = s[i];
            rep(k,0,(n-1))
            {
                bool f = (bool)((1<<k)&(i));
                if(!f)
                    continue;
                rep(l,0,m-1)
                    x[k][l] = '.';
            }

            rep(k,0,(m-1))
            {
                bool f = (bool)((1<<k)&(j));
                if(!f)
                    continue;
                rep(l,0,n-1)
                    x[l][k] = '.';
            }

            int cnt = 0;
            rep(i,0,n-1)
                rep(j,0,m-1)
                {
                    cnt += (x[i][j] == '#');
                }

                      if(cnt == k1)
                ans++;


        }
    cout << ans << endl;
}


signed main()
{
    
    speed;
    int t = 1;
    //cin >> t;
    while(t--)
        solve();
}