Problem: Training is indispensable for achieving good results at ICPC. Rabbit wants to win at ICPC, so he decided to practice today as well.

Today's training is to increase creativity by drawing pictures. Let's draw a pattern well using a square stamp.

I want to use stamps of various sizes to complete the picture of the red, green, and blue streets specified on the 4 x 4 squared paper. The stamp is rectangular and is used to fit the squares. The height and width of the stamp cannot be swapped.

The paper is initially uncolored. When you stamp on paper, the stamped part changes to the color of the stamp, and the color hidden underneath becomes completely invisible. Since the color of the stamp is determined by the ink to be applied, it is possible to choose the color of any stamp. The stamp can be stamped with a part protruding from the paper, and the protruding part is ignored.

It is possible to use one stamp multiple times. You may use the same stamp for different colors. Stamping is a rather nerve-wracking task, so I want to reduce the number of stamps as much as possible.



Input


N
H1 W1
...
HN WN
C1,1C1,2C1,3C1,4
C2,1C2,2C2,3C2,4
C3,1C3,2C3,3C3,4
C4,1C4,2C4,3C4,4


N is the number of stamps, and Hi and Wi (1 ≤ i ≤ N) are integers representing the vertical and horizontal lengths of the i-th stamp, respectively. Ci, j (1 ≤ i ≤ 4, 1 ≤ j ≤ 4) is a character that represents the color of the picture specified for the cells in the i-th row from the top and the j-th column from the left. Red is represented by `R`, green is represented by` G`, and blue is represented by `B`.

Satisfy 1 ≤ N ≤ 16, 1 ≤ Hi ≤ 4, 1 ≤ Wi ≤ 4. The same set as (Hi, Wi) does not appear multiple times.

Output

Print the minimum number of stamps that must be stamped to complete the picture on a single line.

Examples

Input

2
4 4
1 1
RRRR
RRGR
RBRR
RRRR


Output

3


Input

1
2 3
RRGG
BRGG
BRRR
BRRR


Output

5

Solution (in CPP):

Solution:
#include<stdio.h>
#include <iostream>
#include <math.h>
#include <numeric>
#include <vector>
#include <map>
#include <functional>
#include <stdio.h>
#include <array>
#include <algorithm>
#include <string>
#include <string.h>
#include <assert.h>
#include <stdio.h>
#include <queue>
#include<iomanip>
#include<bitset>
#include<stack>
#include<set>
#include<limits>
#include <complex>
using namespace std;


const int mod = 1000000007;
struct Mod {
public:
	int num;
	Mod() : num(0) { ; }
	Mod(long long int n) : num((n % mod + mod) % mod) { ; }
	Mod(int n) : num((n % mod + mod) % mod) { ; }
	operator int() { return num; }
};

Mod operator+(const Mod a, const Mod b) { return Mod((a.num + b.num) % mod); }
Mod operator+(const long long int a, const Mod b) { return Mod(a) + b; }
Mod operator++(Mod &a) { return a + Mod(1); }
Mod operator-(const Mod a, const Mod b) { return Mod((mod + a.num - b.num) % mod); }
Mod operator-(const long long int a, const Mod b) { return Mod(a) - b; }
Mod operator--(Mod &a) { return a - Mod(1); }
Mod operator*(const Mod a, const Mod b) { return Mod(((long long)a.num * b.num) % mod); }
Mod operator*(const long long int a, const Mod b) { return Mod(a)*b; }
Mod operator+=(Mod &a, const Mod b) { return a = a + b; }
Mod operator+=(long long int &a, const Mod b) { return a = a + b; }
Mod operator-=(Mod &a, const Mod b) { return a = a - b; }
Mod operator-=(long long int &a, const Mod b) { return a = a - b; }
Mod operator*=(Mod &a, const Mod b) { return a = a * b; }
Mod operator*=(long long int &a, const Mod b) { return a = a * b; }
Mod operator^(const Mod a, const int n) {
	if (n == 0) return Mod(1);
	Mod res = (a * a) ^ (n / 2);
	if (n % 2) res = res * a;
	return res;
}
Mod inv(const Mod a) { return a ^ (mod - 2); }
Mod operator/(const Mod a, const Mod b) {
	assert(b.num != 0);
	return a * inv(b);
}
Mod operator/(const long long int a, const Mod b) {
	assert(b.num != 0);
	return Mod(a) * inv(b);
}
Mod operator/=(Mod &a, const Mod b) {
	assert(b.num != 0);
	return a = a * inv(b);
}

#define MAX_N 1024000

Mod fact[MAX_N], factinv[MAX_N];
void init() {
	fact[0] = Mod(1); factinv[0] = 1;
	for (int i = 0; i < MAX_N - 1; ++i) {
		fact[i + 1] = fact[i] * Mod(i + 1);
		factinv[i + 1] = factinv[i] / Mod(i + 1);
	}
}
Mod comb(const int a, const int b) {
	return fact[a] * factinv[b] * factinv[a - b];
}


template<typename T>
vector<vector<T>> keisann(const vector<vector<T>>l, const vector<vector<T>>r) {
	vector<vector<T>>ans(l.size(), vector<T>(r[0].size()));
	assert(l[0].size() == r.size());
	for (unsigned int h = 0; h < l.size(); ++h) {
		for (unsigned int i = 0; i < r.size(); ++i) {
			for (unsigned int w = 0; w < r[0].size(); ++w) {

				ans[h][w] += l[h][i] * r[i][w];
			}
		}
	}
	return ans;
}
template<typename T>
vector<vector<T>>powgyou(vector<vector<T>>a, const long long int n) {
	assert(a.size() == a[0].size());
	if (!n) {
		vector<vector<T>>e(a.size(), vector<T>(a[0].size()));
		for (unsigned int i = 0; i < a.size(); ++i) {
			e[i][i] = 1;
		}
		return e;
	}
	if (n == 1)return a;
	else {
		vector<vector<T>>ans(a.size(), vector<T>(a[0].size(), 0));
		ans = powgyou(a, n / 2);
		ans = keisann(ans, ans);
		if (n % 2) {
			ans = keisann(ans, a);
		}
		return ans;
	}
}

#define REP(i,n) for(int i=0;i<(int)n;++i)
#define FOR(i,c) for(auto i=(c).begin();i!=(c).end();++i)
#define ALL(c) (c).begin(), (c).end()

const int INF = 2147483647;
const long long int L_INF = 9223372036854775807;

typedef int Weight;
struct Edge {
	int src, dst;
	Weight weight;
	Edge(int src, int dst, Weight weight) :
		src(src), dst(dst), weight(weight) { }
};
bool operator < (const Edge &e, const Edge &f) {
	return e.weight != f.weight ? e.weight > f.weight : // !!INVERSE!!
	e.src != f.src ? e.src < f.src : e.dst < f.dst;
}
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

typedef vector<Weight> Array;
typedef vector<Array> Matrix;
#define RESIDUE(s,t) (capacity[s][t]-flow[s][t])

vector<pair<int, int>>uses;

//流量0の逆辺も張らないと正しく求まらないので注意
Weight maximumFlow(const Graph &ag, int s, int t) {

	Graph g(ag);
	for (int i = 0; i < ag.size(); ++i) {
		for (int j = 0; j < ag[i].size(); ++j) {
			int d = ag[i][j].dst;
			int s = ag[i][j].src;
			
			bool ok = false;
			for (int k = 0; k < ag[d].size(); ++k) {
				if (ag[d][k].src == s) {
ok = true;
break;
				}
			}
			if (!ok) {
				g[d].push_back(Edge(d, s, 0));
			}
		}
	}

	int n = g.size();
	Matrix flow(n, Array(n)), capacity(n, Array(n));
	REP(u, n) FOR(e, g[u]) capacity[e->src][e->dst] += e->weight;

	Weight total = 0;
	while (1) {
		queue<int> Q; Q.push(s);
		vector<int> prev(n, -1); prev[s] = s;
		while (!Q.empty() && prev[t] < 0) {
			int u = Q.front(); Q.pop();
			FOR(e, g[u]) if (prev[e->dst] < 0 && RESIDUE(u, e->dst) > 0) {
				prev[e->dst] = u;
				Q.push(e->dst);
			}
		}
		if (prev[t] < 0) return total; // prev[x] == -1 <=> t-side
		Weight inc = INF;
		for (int j = t; prev[j] != j; j = prev[j])
			inc = min(inc, RESIDUE(prev[j], j));
		for (int j = t; prev[j] != j; j = prev[j])
			flow[prev[j]][j] += inc, flow[j][prev[j]] -= inc;
		total += inc;
	}
}

map<long long int, int>soinnsuu(long long int a) {
	map<long long int, int>ans;
	for (long long i = 2; i*i <= a; ++i) {
		while (a%i == 0) {
			ans[i]++;
			a /= i;
		}
	}
	if (a != 1)ans[a]++;


	return ans;
}

bool isso(long long int a) {
	if (a == 1)return false;
	for (long long int i = 2; i*i <= a; ++i) {
		if (!(a%i)) {
			return false;
		}
		else {
		}
	}
	return true;
}


struct vert {
	int deg;
	int s;
	int id;
	vert(int a, int b, int c) {
		deg = a;
		s = b;
		id = c;
	}
};


#include<stdio.h>
#include <iostream>
#include <math.h>
#include <numeric>
#include <vector>
#include <map>
#include <functional>
#include <stdio.h>
#include <array>
#include <algorithm>
#include <string>
#include <string.h>
#include <assert.h>
#include <stdio.h>
#include <queue>
#include<iomanip>
#include<bitset>
#include<stack>
#include<set>
#include<limits>
#include <complex>
using namespace std;
class Compare {
public:
	bool operator()(const pair<int, int>&l, const pair<int, int>&r) {
		return l.second > r.second;
	}
};
vector<pair<int, int>>stamps;
int dp[1 << 16];
vector<vector<int>>puz(4, vector<int>(4));;

int getans(const bitset<16>bt) {
	if (dp[bt.to_ulong()] >= 0)return dp[bt.to_ulong()];
	else {
		if (bt.count() == 16)return dp[bt.to_ulong()] = 0;
		else {

			int ans = 16 - bt.count();
			for (int i = 0; i < stamps.size(); ++i) {
				int h = stamps[i].first;
				int w = stamps[i].second;
				for (int l = 1-w; l < 4; ++l) {
					for (int u = 1-h; u < 4; ++u) {
						for (int c = 0; c < 3; ++c) {

							bool ok = true;
							for (int x = max(0,l); x < l + w; ++x) {
								if (x >= 4)break;
								for (int y = max(0,u); y < u + h; ++y) {
									if (y >= 4)break;
									if (bt[y * 4 + x] || puz[y][x] == c) {

									}
									else {
										ok = false;
										goto a;
									}
								}
							}
							if (ok) {

								bitset<16>copybt(bt);
								ok = false;
								for (int x = max(0, l); x < l + w; ++x) {
									if (x >= 4)break;
									for (int y = max(0, u); y < u + h; ++y) {
										if (y >= 4)break;
										if (!copybt[y * 4 + x]) {
											copybt[y * 4 + x] = true;
											ok = true;
										}
									}
								}
								if (ok) {
									ans = min(ans, 1+getans(copybt));
								}
							}
						a:int n;
						}
					}
				}
			}
			return dp[bt.to_ulong()] = ans;
		}
	}
}

int main() {
	memset(dp, -1, sizeof(dp));
	int N; cin >> N;
	for (int i = 0; i < N; ++i) {
		int h, w; cin >> h >> w;
		stamps.push_back(make_pair(h, w));
	}
	for (int i = 0; i < 4; ++i) {
		string st; cin >> st;
		for (int j = 0; j < 4; ++j) {
			if (st[j] == 'R')puz[i][j] = 0;
			else if (st[j] == 'B')puz[i][j] = 1;
			else puz[i][j] = 2;
		}
	}
	int ans = getans(bitset<16>(0));
	cout << ans << endl;
	return 0;
}