#include <bits/stdc++.h>
using namespace std;

using VS = vector<string>;    using LL = long long;
using VI = vector<int>;       using VVI = vector<VI>;
using PII = pair<int, int>;   using PLL = pair<LL, LL>;
using VL = vector<LL>;        using VVL = vector<VL>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))
#define FOR(i, s, e) for (int(i) = (s); (i) < (e); (i)++)
#define FORR(i, s, e) for (int(i) = (s); (i) > (e); (i)--)
#define debug(x) cerr << #x << ": " << x << endl
const int INF = 1e9;                          const LL LINF = 1e16;
const LL MOD = 1000000007;                    const double PI = acos(-1.0);
int DX[8] = { 0, 0, 1, -1, 1, 1, -1, -1 };    int DY[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };

/* -----  2018/09/30  Problem: yukicoder 309  / Link: http://yukicoder.me/problems/no/309  ----- */
/* ------問題------



-----問題ここまで----- */
/* -----解説等-----



----解説ここまで---- */

LL ans = 0LL;

int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);

	int H, W; cin >> H >> W;
	vector<vector<double>>P(H, vector<double>(W));
	FOR(i, 0, H) {
		FOR(j, 0, W) {
			cin >> P[i][j];
			P[i][j] /= 100.0;
		}
	}
	VVI S(H, VI(W));
	FOR(i, 0, H) {
		FOR(j, 0, W) {
			cin >> S[i][j];
		}
	}

	vector<vector<double>>dp(H + 1, vector<double>(1 << W, 0));

	dp[0][0] = 1;
	FOR(i, 0, H) {
		FOR(state, 0, 1 << W) {
			double p = 1;
			FOR(j, 0, W) {
				if (state & 1 << j) {
					p *= P[i][j];
				}
				else {
					p *= 1 - P[i][j];
				}
			}
			// 現在の列がstateであるような確率pは分かっている

			auto ok = [](int s, int k) {
				return s & 1 << k;
			};
			FOR(pre, 0, 1 << W) {
				if (dp[i][pre] < 1e-12)continue;
				int upstate = 0;
				FOR(j, 0, W) {
					if (S[i][j] == 0 && ok(state, j))upstate |= 1 << j;
					if (S[i][j] == 1 && ok(state, j) && ok(pre, j))upstate |= 1 << j;
				}
				FOR(j, 1, W) {
					if (S[i][j] == 1 && ok(state, j) && ok(upstate, j - 1))upstate |= 1 << j;
					if (S[i][j] == 2 && ok(state, j) && ok(pre, j) && ok(upstate, j - 1))upstate |= 1 << j;
				}

				FORR(j, W - 2, 0 - 1) {
					if (S[i][j] == 1 && ok(state, j) && ok(upstate, j + 1))upstate |= 1 << j;
					if (S[i][j] == 2 && ok(state, j) && ok(pre, j) && ok(upstate, j + 1))upstate |= 1 << j;
				}
				FORR(j, W - 2, 0) {
					if (S[i][j] == 2 && ok(state, j) && ok(upstate,j-1) && ok(upstate, j + 1))upstate |= 1 << j;
					if (S[i][j] == 3 && ok(state, j) && ok(pre, j) && ok(upstate, j - 1) && ok(upstate, j + 1))upstate |= 1 << j;
				}


				dp[i + 1][upstate] += p * dp[i][pre];
			}
		}
	}
	double ans = 0;

	FOR(i, 0, H) {
		FOR(state, 0, 1 << W) {
			ans += __builtin_popcount(state)*dp[i + 1][state];
		}
	}

	cout << fixed << setprecision(10) << ans << "\n";

	return 0;
}