Problem:
Score : 1500 points

Problem Statement

We have a grid with 3 rows and N columns.
The cell at the i-th row and j-th column is denoted (i, j).
Initially, each cell (i, j) contains the integer i+3j-3.

A grid with N=5 columns

Snuke can perform the following operation any number of times:

Choose a 3×3 subrectangle of the grid. The placement of integers within the subrectangle is now rotated by 180°.

An example sequence of operations (each chosen subrectangle is colored blue)

Snuke's objective is to manipulate the grid so that each cell (i, j) contains the integer a_{i,j}.
Determine whether it is achievable.

Constraints

5≤N≤10^5

1≤a_{i,j}≤3N

All a_{i,j} are distinct.

Input

The input is given from Standard Input in the following format:

N
a_{1,1} a_{1,2} ... a_{1,N}
a_{2,1} a_{2,2} ... a_{2,N}
a_{3,1} a_{3,2} ... a_{3,N}

Output

If Snuke's objective is achievable, print Yes. Otherwise, print No.

Sample Input 1

5
9 6 15 12 1
8 5 14 11 2
7 4 13 10 3

Sample Output 1

Yes

This case corresponds to the figure in the problem statement.

Sample Input 2

5
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15

Sample Output 2

No

Sample Input 3

5
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15

Sample Output 3

Yes

The objective is already achieved with the initial placement of the integers.

Sample Input 4

6
15 10 3 4 9 16
14 11 2 5 8 17
13 12 1 6 7 18

Sample Output 4

Yes

Sample Input 5

7
21 12 1 16 13 6 7
20 11 2 17 14 5 8
19 10 3 18 15 4 9

Sample Output 5

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define Return(ans) { cout << (ans) << endl; return 0; }
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod

struct SegmentTree
{
    using T = int;

    int N;
    vector<T> dat;

    T id = 0;
    T F(T &a, T &b) { return a + b; }
    
    SegmentTree(int n){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T>(2 * N - 1, id);
    }

    SegmentTree(int n, vector<T> &v){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T>(2 * N - 1, id);
        for(int i = 0; i < n; i++) dat[i + N - 1] = v[i];
        for(int i = N - 2; i >= 0; i--) dat[i] = F(dat[i * 2 + 1], dat[i * 2 + 2]); 
    }

    SegmentTree(){}


    void update(int k, T a){
        k += N - 1;
        dat[k] = a;
        while(k > 0){
            k = (k - 1) / 2;
            dat[k] = F(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }

    void reset() { fill(dat.begin(), dat.end(), id); }

    T get(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id;
        if(a <= l && r <= b) return dat[k];
        else{
            T vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
            T vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
            return F(vl, vr);
        }
    }
    T get(int a, int b) { return get(a, b, 0, 0, N); }

    T val(int k){ return dat[k + N - 1]; }
};

signed main(){

    int N; cin >> N;
    assert(N >= 6);
    mat a(3, vec(N)); cin >> a;

    vec b(N);
    bools flip(N);
    
    REP(i, N){
        REP(j, 3) a[j][i]--;
        int x = (a[1][i] - 1) / 3;

        if(a[0][i] == 3 * x && a[1][i] == 3 * x + 1 && a[2][i] == 3 * x + 2) flip[i] = false;
        else if(a[2][i] == 3 * x && a[1][i] == 3 * x + 1 && a[0][i] == 3 * x + 2) flip[i] = true;
        else{
            Out("No");
            return 0;
        }
        b[i] = x;
        if(b[i] % 2 != i % 2){
            Out("No");
            return 0;
        } 
    }

    vec n(2, 0), m(2, 0);
    REP(k, 2){
        SegmentTree ST(N);
        for(int i = k; i < N; i += 2){
            n[k] += ST.get(b[i], N);
            ST.update(b[i], 1);
            if(flip[i]) m[k]++;
        }
        n[k] %= 2; m[k] %= 2;
    }

    Ans(n[0] == m[1] && n[1] == m[0], "Yes", "No");

    return 0;
}