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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>

#define SIZE 100005
#define PI acosl(-1) //3.14159265358979323846264338327950L 
#define rep(i, N) for (i = 0; i < N; i++) //制御変数iを用いてN回転
#define array(N, t) (t*)calloc(N, sizeof(t)) //t型N要素の1次元配列を動的確保後0クリア(freeを忘れずに)
#define zero(a); {int iter; rep(iter, sizeof(a)/sizeof((a)[0])) (a)[iter]=0;} //1次元配列aのゼロ初期化?
#define Sort(a, t) qsort(a, sizeof(a)/sizeof((a)[0]), sizeof(t), comp) //compを使って1次元配列をソート

typedef long long ll; typedef long double ld;
typedef struct node { int data; struct node *next; } Node;

Node *list[SIZE];
int queue[SIZE]; int q_head = 0, q_tail = 0;

void print_log(); //とりあえず文字を出力してどこまで実行できているか確認(for debug)
void printa_int(int *a, int size); //int型の1次元配列aを出力(for debug)
void printl_int(int size); //リンクリストを出力(for debug)
int comp(const void *a, const void *b); //昇順並び替えに使う比較関数
Node* add_node(int i, int data); void free_list(Node* head);
int queue_empty(); int queue_full();
void enqueue(int data); int dequeue();
ll* factorize(ll N); //Nを素因数分解、得られる配列は素因数,指数,素因数,…となっている(添字はintで指定する必要あり?)
void check_factorization(ll *fact); //素因数分解を表示(for debug)

int H, W, K;
void printhw(int c[], int h, int w) {
    int i, j;
    rep(i, h) {
        rep(j, w) printf(" %d", c[i * w + j]);
        printf("\n");
    }
    printf("\n");
}

int main() {
    int i, j, k, ans = 0;
    scanf("%d %d %d", &H, &W, &K);
    int *C = array(H * W, int);
    int *copy = array(H * W, int);
    rep(i, H) {
        char c;
        scanf("%c", &c);
        rep(j, W) {
            scanf("%c", &c);
            if (c == '#') C[i * W + j] = 1;
        }
    }
    //printhw(C, H, W);
    rep(i, (1 << (H + W))) {
        int count = 0;
        rep(j, H) rep(k, W) {
            if (i >> j & 1) continue;
            if (i >> (H + k) & 1) continue;
            if (C[j * W + k] == 1) count++;
        }
        if (count == K) ans++;
    }
    printf("%d\n", ans);
    return 0;
}

void print_log() {
    printf("=== print log ===\n");
}
void printa_int(int *a, int size) {
    int i;
    printf("[");
    for (i = 0; i < size; i++)
        printf("%d, ", a[i]);
    printf("]\n");
}
void printl_int(int size) {
    int i;
    Node *t;
    rep(i, size) {
        printf("[%d] ", i);
        for (t = list[i]; t; t = t->next) printf("%d, ", t->data);
        printf("\n");
    }
}
int comp(const void *a, const void *b) {
    return *(int*)a - *(int*)b;
}
Node* add_node(int i, int data) {
    if (!list[i]) {
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
    } else {
        Node *t;
        t = list[i];
        list[i] = (Node*)malloc(sizeof(Node*));
        list[i]->data = data;
        list[i]->next = t;
    }
    return list[i];
}
void free_list(Node* head) {
    Node *prev = NULL, *t = head;
    while (t) {
        prev = t;
        t = t->next;
        free(prev);
    }
}
int queue_empty() {
    return q_head == q_tail;
}
int queue_full() {
    return (q_head - 1) % SIZE == q_tail % SIZE;
}
void enqueue(int data) {
    if (queue_full()) {
        printf("full!\n");
        return;
    }
    queue[q_tail++] = data;
    q_tail %= SIZE;
}
int dequeue() {
    if (queue_empty()) {
        printf("empty!\n");
        return -444444;
    }
    int data = queue[q_head++];
    q_head %= SIZE;
    return data;
}
ll* factorize(ll N) {
    ll *res = array(((sqrt(N) + 1) * 2), ll);
    ll i, x = N;
    int count = 0;
    for (i = 2; i * i <= N; i++) {
        ll e = 0;
        while (x % i == 0) {
            x /= i;
            e++;
        }
        if (e > 0) {
            res[count * 2] = i;
            res[count * 2 + 1] = e;
            count++;
        }
    }
    if (x > 1) {
        res[count * 2] = x;
        res[count * 2 + 1] = 1;
    }
    return res;
}
void check_factorization(ll *fact) {
    int i = 1;
    printf("%lld^%lld", fact[0], fact[1]);
    while (fact[i * 2] > 0) {
        printf(" * %lld^%lld", fact[i * 2], fact[i * 2 + 1]);
        i++;
    }
    printf("\n");
}