Given are integer sequences A and B of length 3N. Each of these two sequences contains three copies of each of 1, 2, \dots, N. In other words, A and B are both arrangements of (1, 1, 1, 2, 2, 2, \dots, N, N, N).

Tak can perform the following operation to the sequence A arbitrarily many times:

* Pick a value from 1, 2, \dots, N and call it x. A contains exactly three copies of x. Remove the middle element of these three. After that, append x to the beginning or the end of A.



Check if he can turn A into B. If he can, print the minimum required number of operations to achieve that.

Constraints

* 1 \leq N \leq 33
* A and B are both arrangements of (1, 1, 1, 2, 2, 2, \dots, N, N, N).
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_{3N}
B_1 B_2 ... B_{3N}


Output

If Tak can turn A into B, print the minimum required number of operations to achieve that. Otherwise, print -1.

Examples

Input

3
2 3 1 1 3 2 2 1 3
1 2 2 3 1 2 3 1 3


Output

4


Input

3
1 1 1 2 2 2 3 3 3
1 1 1 2 2 2 3 3 3


Output

0


Input

3
2 3 3 1 1 1 2 2 3
3 2 2 1 1 1 3 3 2


Output

-1


Input

8
3 6 7 5 4 8 4 1 1 3 8 7 3 8 2 4 7 5 2 2 6 5 6 1
7 5 8 1 3 6 7 5 4 8 1 3 3 8 2 4 2 6 5 6 1 4 7 2


Output

7
#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double; 
using str = string; // yay python!

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>; 
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
using vpd = vector<pd>;

#define tcT template<class T
// ^ lol this makes everything look weird but I'll try it
tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 

// pairs
#define mp make_pair
#define f first
#define s second

// vectors
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define sor(x) sort(all(x)) 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

// helper funcs
constexpr int pct(int x) { return __builtin_popcount(x); } // # of bits set
constexpr int bits(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

tcT> bool ckmin(T& a, const T& b) {
	return b < a ? a = b, 1 : 0; } // set a = min(a,b)
tcT> bool ckmax(T& a, const T& b) {
	return a < b ? a = b, 1 : 0; }

#define tcTU tcT, class U
tcTU> T fstTrue(T lo, T hi, U f) {
	hi ++; assert(lo <= hi); // assuming f is increasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo)/2;
		f(mid) ? hi = mid : lo = mid+1; 
	} 
	return lo;
}
tcTU> T lstTrue(T lo, T hi, U f) {
	lo --; assert(lo <= hi); // assuming f is decreasing
	while (lo < hi) { // find first index such that f is true 
		T mid = lo+(hi-lo+1)/2;
		f(mid) ? lo = mid : hi = mid-1;
	} 
	return lo;
}
tcT> void remDup(vector<T>& v) { // sort and remove duplicates
	sort(all(v)); v.erase(unique(all(v)),end(v)); }
tcTU> void erase(T& t, const U& u) { // don't erase
	auto it = t.find(u); assert(it != end(t));
	t.erase(u); } // element that doesn't exist from (multi)set

// INPUT
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(vector<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }

tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(vector<T>& x) { trav(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

// OUTPUT
tcT> void pr(T x) { cout << ts(x); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // print w/ spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU> void DBG(const T& t, const U&... u) {
	cerr << ts(t); if (sizeof...(u)) cerr << ", ";
	DBG(u...); }
#ifdef LOCAL // compile with -DLOCAL, chk -> fake assert
	#define dbg(...) cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
	#define chk(...) if (!(__VA_ARGS__)) cerr << "Line(" << __LINE__ << ") -> function(" \
		 << __FUNCTION__  << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n", exit(0);
#else
	#define dbg(...) 0
	#define chk(...) 0
#endif

// FILE I/O
void setIn(str s) { freopen(s.c_str(),"r",stdin); }
void setOut(str s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

/**
 * Description: Tarjan's, DFS once to generate 
 	* strongly connected components in topological order. $a,b$
 	* in same component if both $a\to b$ and $b\to a$ exist.
 	* Uses less memory than Kosaraju b/c doesn't store reverse edges.
 * Time: O(N+M)
 * Source: KACTL
 	* https://github.com/kth-competitive-programming/kactl/blob/master/content/graph/SCC.h
 * Verification: https://cses.fi/problemset/task/1686/
 */

struct SCC {
	int N, ti = 0; vector<vi> adj;
	vi disc, comp, st, comps;
	void init(int _N) { N = _N; adj.rsz(N), disc.rsz(N), comp = vi(N,-1); }
	void ae(int x, int y) { adj[x].pb(y); }
	int dfs(int x) {
		int low = disc[x] = ++ti; st.pb(x); // disc[y] != 0 -> in stack
		trav(y,adj[x]) if (comp[y] == -1) ckmin(low,disc[y]?:dfs(y)); 
		if (low == disc[x]) { // make new SCC, pop off stack until you find x
			comps.pb(x); for (int y = -1; y != x;) 
				comp[y = st.bk] = x, st.pop_back();
		}
		return low;
	}
	void gen() {
		F0R(i,N) if (!disc[i]) dfs(i);
		reverse(all(comps));
	}
};


/**
 * Description: Calculates a valid assignment to boolean variables a, b, c,... 
 	* to a 2-SAT problem, so that an expression of the type 
 	* $(a\|\|b)\&\&(!a\|\|c)\&\&(d\|\|!b)\&\&...$ becomes true, 
 	* or reports that it is unsatisfiable.
 	* Negated variables are represented by bit-inversions (\texttt{\tilde{}x}).
 * Usage:
	* TwoSat ts;
	* ts.either(0, \tilde3); // Var 0 is true or var 3 is false
	* ts.setVal(2); // Var 2 is true
	* ts.atMostOne({0,\tilde1,2}); // <= 1 of vars 0, \tilde1 and 2 are true
	* ts.solve(N); // Returns true iff it is solvable
	* ts.ans[0..N-1] holds the assigned values to the vars
 * Source: KACTL
 * Verification: https://codeforces.com/contest/1007/problem/D
 	* https://cses.fi/problemset/task/1684/
 */

// #include "SCC (12.1).h"

struct TwoSAT {
	int N; SCC S; vb ans;
	void init(int _N) { N = _N; S.init(2*N); ans.rsz(N); }
	int addVar() { return N++; }
	void either(int x, int y) { 
		x = max(2*x,-1-2*x), y = max(2*y,-1-2*y);
		S.ae(x^1,y); S.ae(y^1,x); }
	void implies(int x, int y) { either(~x,y); }
	void must(int x) { either(x,x); }
	void atMostOne(vi li) {
		if (sz(li) <= 1) return;
		int cur = ~li[0];
		FOR(i,2,sz(li)) {
			int next = addVar();
			either(cur,~li[i]); either(cur,next);
			either(~li[i],next); cur = ~next;
		}
		either(cur,~li[1]);
	}
	bool solve() {
		// if (_N != -1) N = _N, S.init(2*N);
		S.gen(); reverse(all(S.comps)); // reverse topological order
		for (int i = 0; i < 2*N; i += 2) 
			if (S.comp[i] == S.comp[i^1]) return 0;
		vi tmp(2*N); trav(i,S.comps) if (!tmp[i]) 
			tmp[i] = 1, tmp[S.comp[i^1]] = -1;
		F0R(i,N) if (tmp[S.comp[2*i]] == 1) ans[i] = 1;
		return 1;
	}
};

int N;
vi A,B;

V<vi> cat(vi v) {
	map<int,vi> todo;
	F0R(i,sz(v)) todo[v[i]].pb(i);
	V<vi> ans;
	trav(t,todo) ans.pb(t.s);
	return ans;
}

V<vi> apos,bpos;

int val(int l, int r) {
	dbg("DOING",l,r);
	vpi match;
	int ms = 0;
	vpi typ1, typ2;
	vi cand;
	F0R(z,N) {
		str cur;
		trav(t,bpos[z]) {
			if (t < l) {
				cur += 'L';
			} else if (t < r) {
				cur += 'M';
			} else {
				cur += 'R';
			}
		}
		dbg(z,cur);
		trav(t,cur) ms += t == 'M';
		if (cur == "LLM") {
			match.pb({apos[z][2],bpos[z][2]});
		} else if (cur == "LLR") { // type 1
			typ1.pb({bpos[z][0],bpos[z][2]});
		} else if (cur == "LMM") { //
			match.pb({apos[z][0],bpos[z][1]});
			match.pb({apos[z][2],bpos[z][2]});
		} else if (cur == "LMR") { // type 1 or type 2
			cand.pb(z);
		} else if (cur == "LRR") { // type 2
			typ2.pb({bpos[z][0],bpos[z][2]});
		} else if (cur == "MMM") { 
			F0R(i,3) match.pb({apos[z][i],bpos[z][i]});
		} else if (cur == "MMR") {
			match.pb({apos[z][0],bpos[z][0]});
			match.pb({apos[z][2],bpos[z][1]});
		} else if (cur == "MRR") {
			match.pb({apos[z][0],bpos[z][0]});
		} else {
			return MOD;
		}
	}
	sort(all(match));
	FOR(i,1,sz(match)) if (match[i].s < match[i-1].s) return MOD;
	dbg(typ1,typ2);
	trav(a,typ1) trav(b,typ2) {
		if (b.f < a.f && b.s < a.s) return MOD;
	}
	TwoSAT T; T.init(N);
	auto incomparable = [](pi a, pi b) {
		return (a.f > b.f)!= (a.s > b.s);
	};
	// https://docs.google.com/document/d/1DdleBCMgQEO5thXH5_F75Zk93hzsJsEgWUMLYp5s998/edit
	auto badMatch = [&](pi p) {
		trav(t,match) {
			if (incomparable(t,p)) return 1;
		}
		return 0;
	};
	trav(z,cand) {
		{
			pi a = {bpos[z][0],bpos[z][2]};
			trav(b,typ2) if (b.f < a.f && b.s < a.s) {
				T.either(~z,~z);
				// dbg("BAD1");
			}
			if (badMatch({apos[z][0],bpos[z][1]})) {
				T.either(~z,~z);
				// dbg("BAD1");
			}
			// typ1.pb({bpos[z][0],bpos[z][2]}), match.pb({apos[z][0],bpos[z][1]});
		}
		{
			pi b = {bpos[z][0],bpos[z][2]};
			trav(a,typ1) if (b.f < a.f && b.s < a.s) {
				T.either(z,z);
				// dbg("BAD2");
			}
			if (badMatch({apos[z][2],bpos[z][1]})) {
				T.either(z,z);
				// dbg("BAD2");
			}
			// typ2.pb({bpos[z][0],bpos[z][2]}), match.pb({apos[z][2],bpos[z][1]});
		}
	}
	trav(x,cand) trav(y,cand) if (x != y) {
		F0R(X,2) F0R(Y,2) {
			int xWut = x; if (X) xWut = ~xWut;
			int yWut = y; if (Y) yWut = ~yWut;
			pi a = {apos[x][2*X],bpos[x][1]}, b = {apos[y][2*Y],bpos[y][1]};
			if (incomparable(a,b)) T.either(~xWut,~yWut);
		}
	}
	trav(x,cand) trav(y,cand) if (x != y) {
		pi a = {bpos[x][0],bpos[x][2]};
		pi b = {bpos[y][0],bpos[y][2]};
		if (b.f < a.f && b.s < a.s) T.either(~x,y);
		// typ1.pb({bpos[z][0],bpos[z][2]}), match.pb({apos[z][0],bpos[z][1]});
		// typ2.pb({bpos[z][0],bpos[z][2]}), match.pb({apos[z][2],bpos[z][1]});
	}
	if (!T.solve()) return MOD;
	// dbg("COOL",l,r);
	return 3*N-ms;
}

int main() {
	setIO(); re(N); A.rsz(3*N), B.rsz(3*N); re(A,B);
	apos = cat(A), bpos = cat(B);
	int ans = MOD;
	F0R(l,3*N+1) FOR(r,l,3*N+1) ckmin(ans,val(l,r));
	if (ans == MOD) ans = -1;
	ps(ans);
	// you should actually read the stuff at the bottom
}

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
	* DON'T GET STUCK ON ONE APPROACH
*/
