//dp[i][j]表示前i个小朋友中左括号与右括号的差为j的最大实力值
//dp1[i][j]表示前i个小朋友中左括号与右括号的差为j的方案总数。
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
const int maxn = 1e7 + 5;
const int mod = 998244353;
int n, w[maxn], c[maxn];
long long dp[5][maxn], dp1[5][maxn];
int s = 0;

void freopen() { 
	freopen("pick.in", "r", stdin);
	freopen("pick.out", "w", stdout);
}

int MAX(int x, int y) { 
	return x > y ? x : y;
}

int main() { 
	freopen();
	scanf("%d",&n);
	for(int i = 1;i <= n;i ++) { 
		scanf("%d", &w[i]);
	}
	for(int i = 1;i <= n;i ++) { 
		scanf("%d", &c[i]);
	}
	dp[0][0] = 0, s = 0;
	for(int i = 1;i <= n;i ++){ 
		s = 1 - s;
		for(int j = 0;j <= n;j ++){ 
			if(!j) { 
				dp[s][j] = max(dp[1 - s][j] , dp[1 - s][j + 1]);
				if(j >= c[i]) dp[s][j] = max(dp[s][j], dp[1 - s][j - c[i]] + w[i]);
			}
			else { 
				dp[s][j] = dp[1 - s][j + 1];
				if(j >= c[i]) dp[s][j] = max(dp[s][j], dp[1 - s][j - c[i]] + w[i]);
			}
		}
	}
	printf("%lld ",dp[s][0]);
	dp1[0][0] = 1, s = 0;
	for(int i = 1;i <= n;i ++){ 
		s = 1  - s;
		for(int j = 0;j <= n;j ++){ 
			if(!j){ 
				dp1[s][j] = (dp1[1 - s][j] + dp1[1 - s][j + 1]) % mod;
				if(j >= c[i]) dp1[s][j] = (dp1[s][j] + dp1[1 - s][j - c[i]]) % mod;
			}else{ 
				dp1[s][j] = dp1[1 - s][j + 1];
				if(j >= c[i]) dp1[s][j] = (dp1[s][j] + dp1[1 - s][j - c[i]]) % mod;
			}
		}
	}
	printf("%lld", dp1[s][0]);
	return 0;
}