题目链接：

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Have you ever played Hanabi? If not, then you've got to try it out! This problem deals with a simplified version of the game.

Overall, the game has 25 types of cards (5 distinct colors and 5 distinct values). Borya is holding n cards. The game is somewhat complicated by the fact that everybody sees Borya's cards except for Borya himself. Borya knows which cards he has but he knows nothing about the order they lie in. Note that Borya can have multiple identical cards (and for each of the 25 types of cards he knows exactly how many cards of this type he has).

The aim of the other players is to achieve the state when Borya knows the color and number value of each of his cards. For that, other players can give him hints. The hints can be of two types: color hints and value hints.

A color hint goes like that: a player names some color and points at all the cards of this color.

Similarly goes the value hint. A player names some value and points at all the cards that contain the value.

Determine what minimum number of hints the other players should make for Borya to be certain about each card's color and value.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of Borya's cards. The next line contains the descriptions of n cards. The description of each card consists of exactly two characters. The first character shows the color (overall this position can contain five distinct letters — R, G, B, Y, W). The second character shows the card's value (a digit from 1 to 5). Borya doesn't know exact order of the cards they lie in.

Output

Print a single integer — the minimum number of hints that the other players should make.

Examples

input

2 G3 G3

output

0

input

4 G4 R4 R3 B3

output

2

input

5 B1 Y1 W1 G1 R1

output

4 题意： 现在知道有哪些牌,但不知具体顺序,每次询问都会知道这个询问的相关的牌的位置,问最少需要多少次询问; 思路： 变成二维的点,划线后被两次覆盖的点可以删去,剩下的一行或者一列只有一个的也可以删去了; 看最后如果剩下的个数<=1就这种选择就可行; AC代码：

#include 
        
         using namespace std;typedef long long LL;const int maxn=5e5+10;int n,a[6][6],x[6],y[6],b[6][6],numx[6],numy[6];char s[2];int check(int temp){    for(int i=0;i<5;i++)        for(int j=0;j<5;j++)            b[i][j]=a[i][j];    for(int i=0;i<5;i++)if((temp>>i)&1)x[i]=1;else x[i]=0;    for(int i=5;i<10;i++)if((temp>>i)&1)y[i-5]=1;else y[i-5]=0;    for(int i=0;i<5;i++)    {        if(x[i])        {            int num=0;            for(int j=0;j<5;j++)if(b[i][j])num++;            if(num==1)            {                for(int j=0;j<5;j++)b[i][j]=0;            }        }    }    for(int i=0;i<5;i++)    {        if(y[i])        {            for(int j=0;j<5;j++)            {                if(b[j][i]&&x[j])b[j][i]=0;            }            int num=0;            for(int j=0;j<5;j++)if(b[j][i])num++;            if(num==1)            {                for(int j=0;j<5;j++)b[j][i]=0;            }        }    }    for(int i=0;i<5;i++)    {        if(x[i])        {            int num=0;            for(int j=0;j<5;j++)if(b[i][j])num++;            if(num==1)            {                for(int j=0;j<5;j++)b[i][j]=0;            }        }    }    int num=0;    for(int i=0;i<5;i++)        for(int j=0;j<5;j++)            if(b[i][j])num++;    if(num>1)return 10;    int ans=0;//cout<
         
          <<"*****"<
          
           >=1;}    return ans;}int main(){    scanf("%d",&n);    int fx,fy;    for(int i=1;i<=n;i++)    {        scanf("%s",s);        if(s[0]=='R')fx=0;        else if(s[0]=='B')fx=1;        else if(s[0]=='Y')fx=2;        else if(s[0]=='W')fx=3;        else fx=4;        fy=s[1]-'0'-1;        a[fx][fy]++;    }    int num=0;    for(int i=0;i<5;i++)        for(int j=0;j<5;j++)            if(a[i][j])num++;    if(num<=1){printf("0\n");return 0;}    int ans=10;    for(int i=0;i<(1<<10);i++)ans=min(ans,check(i));    cout<
           
            <