#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h" // define macro "/D__MAI"

using namespace std;
typedef long long int ll;

#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:v){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(auto cnt=0ll;(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define MD 1000000007ll
#define PI 3.1415926535897932384626433832795
template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o; }
template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
inline ll rand_range(ll l, ll h) {
    return uniform_int_distribution<ll>(l, h)(randdev);
}

#if defined(_WIN32) || defined(_WIN64)
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#elif defined(__GNUC__)
#else
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0; T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var * sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }
        inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiblechar(cc); cc = getchar_unlocked());
            for (; isvisiblechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
            return *this;
        }
        template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
    };
    class MaiPrinter {
    public:
        template<typename T>
        void output_integer(T var) {
            if (var == 0) { putchar_unlocked('0'); return; }
            if (var < 0)
                putchar_unlocked('-'),
                var = -var;
            char stack[32]; int stack_p = 0;
            while (var)
                stack[stack_p++] = '0' + (var % 10),
                var /= 10;
            while (stack_p)
                putchar_unlocked(stack[--stack_p]);
        }
        inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }
        inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }
        inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }
        inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }
        inline MaiPrinter& operator<<(const string& str) {
            const char* p = str.c_str();
            const char* l = p + str.size();
            while (p < l) putchar_unlocked(*p++);
            return *this;
        }
        template<typename IT> void join(IT begin, IT end, char sep = '\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }
    };
}
MaiScanner scanner;
MaiPrinter printer;


/*
入力をうまく変換すると，

コインの状態vector<boolean>と，操作の集合set<pair<int,int>>が与えられる．
操作(x,y)を実行すると，x番目のコインとy番目のコインを反転させる．(x==yなら，x番目のコインのみを反転させる)
全て表向きにできるか？

コインを頂点，操作を辺と置き換えることができる．すると，

幾つかの辺を選んで，ww[i]=0の時は奇頂点，ww[i]=1の時は偶頂点となるようにしたい．
条件を満たす辺の選び方は存在するか？

一筆書きの話に非常によく似ている

両端点がww[i]==0,他がww[i]!=0となるようなパスが存在するならば，連続でそのパスを選択することで，パスが通る頂点は条件を満たす．

*/


class Graph2d {
public:
    typedef int W_T;
    size_t n;
    vector<W_T> matrix;

    Graph2d(size_t size) :n(size), matrix(size*size) {};

    void resize(size_t s) {
        n = s; matrix.resize(n*n);
    }
    void resize(size_t s, W_T val) {
        n = s; matrix.resize(n*n, val);
    }

    inline W_T& at(int y, int x) { return matrix[y*n + x]; }
    inline W_T& operator()(int y, int x) { return matrix[y*n + x]; }
    inline W_T at(int y, int x) const { return matrix[y*n + x]; }
    inline W_T operator()(int y, int x) const { return matrix[y*n + x]; }

    inline void connect(int u, int v, W_T dist = 1) {
        at(u, v) = at(v, u) = dist;
    }
    inline void connect_d(int from, int to, W_T dist = 1) { // directedEdge u->v
        at(from, to) = dist;
    }
};


void warshall_floyd(Graph2d& g) {
    int i, j, k;
    for (i = 0; i < g.n; i++) {
        for (j = 0; j < g.n; j++) {
            for (k = 0; k < g.n; k++) {
                g(j, k) = min(g(j, k), g(j, i) + g(i, k));
            }
        }
    }
}



ll m, n, kei;

int dd[111];
int ww[111];

vector<int> flip_from[111]; // iを反転すると，

const int inf = 1e9;

int main() {
    scanner >> n;
    scanner.in(dd, dd + n);
    scanner.in(ww, ww + n);

    Graph2d graph(n);
    fill(ALL(graph.matrix), inf);

    repeat(i, n) {
        dd[i] %= n;
        int x = (i + dd[i]) % n;
        int y = (i - dd[i] + n) % n;

        if (x == y) {
            ww[x] = -1;
        }
        else {
            graph.connect(x, y);
        }
    }

    warshall_floyd(graph);

    repeat(i, n) {
        if (ww[i] != 0) continue;
        iterate(j, 0, n) {
            if (i == j) continue;
            if (ww[j] == 1) continue;
            if (graph(i, j) < inf) {
                if (ww[j] == 0)
                    ww[j] = 1;
                ww[i] = 1;
                break;
            }
        }
        if (ww[i] == 0)
            bye("No");
    }

    bye("Yes");

    return 0;
}