A. Infinity Gauntlet

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:

  • the Power Gem of purple color,
  • the Time Gem of green color,
  • the Space Gem of blue color,
  • the Soul Gem of orange color,
  • the Reality Gem of red color,
  • the Mind Gem of yellow color.

Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

Input

In the first line of input there is one integer nn (0≤n≤60≤n≤6) — the number of Gems in Infinity Gauntlet.

In next nn lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.

Output

In the first line output one integer mm (0≤m≤60≤m≤6) — the number of absent Gems.

Then in mm lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.

Examples

input

Copy

4
red
purple
yellow
orange

output

Copy

2
Space
Time

input

Copy

0

output

Copy

6
Time
Mind
Soul
Power
Reality
Space

Note

In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.

In the second sample Thanos doesn't have any Gems, so he needs all six.

 

 

 

灭霸的无限手套上有6颗宝石，每个宝石对应不同的属性：空间，时间，意识，力量，现实，灵魂。。给你其种几个颜色，问剩下的几颗宝石属性是啥。。。（不知道为什么，我总是能把水题做得那么复杂QAQ）

代码如下：

#include<iostream>
#include<map>
#include<set>
#include<string>
using namespace std;
map<string,string> m;
string b[6];
void list_m(){
	m["purple"]="Power";
	m["green"]="Time";         //map,颜色和属性对应
	m["blue"]="Space";
	m["orange"]="Soul";
	m["red"]="Reality";
	m["yellow"]="Mind";
	b[0]="purple";
	b[1]="green";             //string数组，编号和颜色对应
	b[2]="blue";
	b[3]="orange";
	b[4]="red";
	b[5]="yellow";
}
int main(){
	string a[6];
	list_m();
	set<int> vv;         //集合，保存了0~5一共6个数字(编号)
	int n;
	cin>>n;
	int s=6-n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];          //读入已有的几种颜色
		
    }
    if(n==6){
    	cout<<0<<endl;
	}
	else{
	    for(int i=0;i<6;i++)
	     {
	     vv.insert(i);      
		 }
	    int v[6],k=0;
		for(int i=0;i<6;i++){
			for(int j=0;j<n;j++){
				if(a[j]==b[i]){          //取这几个颜色的编号
					v[k++]=i;
				}
			}
		}

		 set<int> ::iterator it;

		 for(it=vv.begin();it!=vv.end();it++)
		 { for(int i=0;i<k;i++)
		   {
		 	if(*it==v[i])
		    {
	 		  vv.erase(it++);//在集合里删除这几个颜色的编号，剩下的就是要输出的颜色
		    }
	       }
	     }

		 cout<<vv.size()<<endl;
		 for(it=vv.begin();it!=vv.end();it++)
		 {
		    cout<<m[b[*it]]<<endl; //输出
		 }
	 }        //看不懂算了，我看着也费劲。……
	return 0;
}

简洁版：我的DEV不支持新标准的auto关键字，哎，换编辑器。

#include<bits/stdc++.h>
using namespace std;
map<string,string> m;
void list_m(){
	m["purple"]="Power";
	m["green"]="Time";         
	m["blue"]="Space";
	m["orange"]="Soul";
	m["red"]="Reality";
	m["yellow"]="Mind";
}
int main(){
	list_m();
	string a;
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a;
		m[a]="";
    }
    cout<<6-n<<endl;
    for(auto t:m){
    	if(t.second != "")
    	cout<< t.second <<endl;
	}
    
	return 0;
}