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

signed main() {
    int n;cin >> n;
    vector<int> a(n + 1), b(n + 1), c(n + 1), lst;
    int sum = 0;
    for(int i = 1; i <= n; ++i) cin >> a[i];
    for(int i = 1; i <= n; ++i) cin >> b[i];
    for(int i = 1; i <= n; ++i) cin >> c[i];
    for(int i = 1; i <= n; ++i) {
        if(a[i]) {
            a[i] += c[i], c[i] = 0;
            int add = min(a[i], b[i]);
            sum += add, a[i] -= add, b[i] -= add;
        }
        else {
            if(sum > 0 and b[i] > 0) {
                a[i] = 1 + c[i], c[i] = 0, sum--;
                int add = min(a[i], b[i]);
                sum += add, a[i] -= add, b[i] -= add;
            }
            else if(sum > 0 and c[i] > 0) lst.emplace_back(c[i]);
        }
    }
    sort(begin(lst), end(lst), greater<int> ());
    for(int i = 1; i <= n; ++i) {
        if(sum) cout << a[i] + c[i] + sum << " ";
        else cout << a[i] << " ";
    }
    cout << "\n";
    for(int i = 1; i <= n; ++i) sum += a[i];
    for(int i = 0; i < min<int>(sum, lst.size()); ++i) {
        sum += lst[i];
    }
    cout << sum << "\n";
    return 0;
}
/*
如果a[i] > 0那么把c[i]先全变成a[i],
如果没有就考虑是否把别的换成a[i],换成i的前提是b[i] != 0,
如果不够的话应该优先给c[i]大的
最后答案就是所有都转化成i,然后用前缀和优化
*/