C++ :

/*Code by Leo2011*/
#include <bits/stdc++.h>

#define INF 0x3f3f3f3f
#define EPS 1e-8
#define FOR(i, l, r) for (ll(i) = (l); (i) <= (r); ++(i))
#define log printf
#define IOS                      \
	ios::sync_with_stdio(false); \
	cin.tie(nullptr);            \
	cout.tie(nullptr);

using namespace std;

typedef __int128 i128;
typedef long long ll;
typedef pair<ll, ll> PII;

const int N = 510, M = 310;
int m, n, t, x, a[N][N], dp[N][N][M], ans;  // dp[i][j][k] = (1, 1) 到(i, j)，已经使用了k次机会的最大值
string s;

template <typename T>

inline T read() {
	T sum = 0, fl = 1;
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar())
		if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
	return sum * fl;
}

template <typename T>

inline void write(T x) {
	if (x < 0) {
		putchar('-'), write<T>(-x);
		return;
	}
	static T sta[35];
	ll top = 0;
	do { sta[top++] = x % 10, x /= 10; } while (x);
	while (top) putchar(sta[--top] + 48);
}

int main() {
    IOS;
    cin >> t;
    while (t--) {
        cin >> n >> m >> x;
        FOR(i, 1, n) {
            cin >> s;
            for (int j = 0;j < m;++j) {
                if (s[j] == '?') a[i][j + 1] = -INF;
                else a[i][j + 1] = s[j] - '0';
            }
        }
        FOR(i, 1, n) FOR(j, 1, m) FOR(k, 0, x) {
              // 要用到 k - 1, 必须正序遍历。
            switch(a[i][j]) {
                case -INF:
                    if (k > 0) dp[i][j][k] = max(dp[i][j - 1][k - 1], dp[i - 1][j][k - 1]) + 1;
                    else dp[i][j][k] = max(dp[i][j - 1][k], dp[i - 1][j][k]);
                    break;
                case 1:
                    dp[i][j][k] = max(dp[i][j - 1][k], dp[i - 1][j][k]) + 1;
                    break;
                case 0:
                    dp[i][j][k] = max(dp[i][j - 1][k], dp[i - 1][j][k]);
                    break;
            }
        }
        FOR(i, 0, x) ans = max(ans, dp[n][m][i]);  // 不超过就是不一定，要遍历一遍
        cout << ans << endl;
        memset(dp, 0, sizeof(dp));
        memset(a, 0, sizeof(a));
        ans = -INF;
        // 多组数据要清空
    }
	return 0;
}