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)

// Author :: <Hitesh_Saini>
#include<bits/stdc++.h>

#define __speed() ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define dbg(x) cout << "--- " << #x << " = " << x << " ---\n";
#define findmax(v) *max_element(v.begin(), v.end())
#define _sum(a) accumulate(a.begin(), a.end(), 0)
#define tt int t;for(cin>>t; t--; )
#define f0(i, n) for(i = 0; i < int(n); i++)
#define f1(i, n) for(i = 1; i<= int(n); i++)
#define all(x) x.begin(), x.end()
#define pq priority_queue
#define EB emplace_back
#define PB push_back
#define endl "\n"
#define S second
#define ar array
#define F first

using namespace std;
using mii = map<int, int>;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
const int mod = 1e9+7,mxN = 2e6+5;

template<typename... Args>void print(Args... args){((cout << args << " "), ...);cout << endl;}
ll mod_pow(ll x, ll y){ll res=1;x%=mod;for(;y>0; y>>=1, x=(x*x)%mod)if(y&1)res=(res*x) % mod; return res; }ll mod_inv ( ll n ){ll ans = mod_pow ( n , mod - 2 ) ;return ans ;}ll mul(ll a, ll b){a%=mod, b%=mod;return (a*b)%mod;}ll sub(ll a,ll b){a%=mod, b%=mod;return (a-b+mod)%mod;}ll add(ll a, ll b){a%=mod, b%=mod;return (a+b)%mod;}
bool prime(ll n){if(n<2)return 0;if(n<4)return 1;if(n%2==0||n%3==0)return 0;for(ll i=5;i*i<=n;i+=6)if(n%i==0||n%(i+2)==0)return 0;return 1;}

/*#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>  
using namespace __gnu_pbds; 
#define iset tree<ll , null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define iof order_of_key
#define findat find_by_order */
int bcr[7], bcc[7], total, tbcr[7], tbcc[7];
string st[7];
void solve(){ //B
	int i, j, p, q, r, c, k, ans=0, y;
	char x;
	cin >> r >> c >> k;
	f0(i, r)
		cin >> st[i];
		f0(p, (1<<r))
			f0(q, (1<<c)){
				int bc =0;
				f0(i, r)
					f0(j, c)
						if(((p >> i) & 1) == 0 && ((q >> j) & 1) == 0 && st[i][j] == '#')
							bc++;
				ans += bc==k;
			}
	print(ans);
}

int main(){
	#ifndef ONLINE_JUDGE
		freopen("input.txt", "r", stdin);
		freopen("output.txt", "w", stdout);
	#endif
    __speed();
    solve();
    return 0;
}