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 n

(0≤n≤6

) — the number of Gems in Infinity Gauntlet.

In next n

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 m

(0≤m≤6

) — the number of absent Gems.

Then in m

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.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 900005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
map<string, string>mp;
map<string, int>mp2;
string ss[10];
int main() {
//ios::sync_with_stdio(0);
mp["purple"] = "Power"; mp["green"] = "Time";
mp["blue"] = "Space"; mp["orange"] = "Soul";
mp["red"] = "Reality"; mp["yellow"] = "Mind";
rdint(n);
ss[1] = "red"; ss[2] = "blue"; ss[3] = "yellow";
ss[4] = "purple"; ss[5] = "green"; ss[6] = "orange";
for (int i = 1; i <= n; i++) {
string tmp; cin >> tmp;
mp2[tmp] = 1;
}
cout << 6 - n << endl;
for (int i = 1; i <= 6; i++) {
if (!mp2[ss[i]]) {
cout << mp[ss[i]] << endl;
}
}
return 0;
}

CF987A Infinity Gauntlet 模拟的更多相关文章

  1. CF987A Infinity Gauntlet【STL】

    [链接]:CF987A [分析]:运用map [代码]: #include <iostream> #include<queue> #include<string.h> ...

  2. Codeforces Round #485 (Div. 2) A. Infinity Gauntlet

    Codeforces Round #485 (Div. 2) A. Infinity Gauntlet 题目连接: http://codeforces.com/contest/987/problem/ ...

  3. Codeforces 987A. Infinity Gauntlet(手速题,map存一下输出即可)

    解法: 1.先将对应的字符串存入map. 2.然后将输入的串的second置为空. 3.输出6-n,输出map中的非空串. 代码: #include <bits/stdc++.h> usi ...

  4. April Fools Day Contest 2019 A. Thanos Sort

    A. Thanos Sort time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  5. April Fools Day Contest 2019: editorial回顾补题

    A. Thanos Sort time limit per test 1 second memory limit per test 256 megabytes input standard input ...

  6. EZ 2018 06 17 NOIP2018 模拟赛(十九)

    这次的题目难得的水,但是由于许多哲学的原因,第二题题意表述很迷. 然后是真的猜题意了搞了. 不过这样都可以涨Rating我也是服了. Upt:链接莫名又消失了 A. 「NOIP2017模拟赛11.03 ...

  7. 2017NOIP模拟赛-科普基地

    今天回来打的第一场NOIP难度的试题,结果惨不忍睹.写一下每道题的做法,然后每道题犯的__弱智__错误 UPD:2018.9.15 突然这篇题解就变成很多大佬要看的了,因为之前是写给自己看的,所以写的 ...

  8. CF--思维练习--CodeForces - 216C - Hiring Staff (思维+模拟)

    ACM思维题训练集合 A new Berland businessman Vitaly is going to open a household appliances' store. All he's ...

  9. App开发:模拟服务器数据接口 - MockApi

    为了方便app开发过程中,不受服务器接口的限制,便于客户端功能的快速测试,可以在客户端实现一个模拟服务器数据接口的MockApi模块.本篇文章就尝试为使用gradle的android项目设计实现Moc ...

随机推荐

  1. spring学习-1

    spring框架的组件结构图 IOC(Inversion of Control):反转控制,思想是反转资源获取方向,传统的资源查找方向是组件向容器发起请求资源查找,作为回应,容器适时的返回资源,IOC ...

  2. 搭建 Http Dynamic Streaming 点播/直播服务器

    1.        HTTP Origin Module的处理数据流: a)         客户端发送媒体索引请求到Apache.例如: http://www.example.com/media/  ...

  3. ACM学习历程—HDU5478 Can you find it(数论)(2015上海网赛11题)

    Problem Description Given a prime number C(1≤C≤2×105), and three integers k1, b1, k2 (1≤k1,k2,b1≤109 ...

  4. CentOS6 下Samba服务器的安装与配置

    原地址:http://www.cnblogs.com/mchina/archive/2012/12/18/2816717.html 一.简介 Samba是一个能让Linux系统应用Microsoft网 ...

  5. keepalive安装和配置

    1.下载安装包并解压 sudo wget http://www.keepalived.org/software/keepalived-1.2.13.tar.gz tar zxvf keepalived ...

  6. 优秀开源项目之三:高性能、高并发、高扩展性和可读性的网络服务器架构State Threads

    译文在后面. State Threads for Internet Applications Introduction State Threads is an application library ...

  7. MySQL(介绍1)

    数据库(Database)是按照数据结构来组织.存储和管理数据的仓库: 也可以将数据存储在文件中,但是在文件中读写数据速度相对较慢. 在WEB应用方面MySQL是最好的RDBMS(Relational ...

  8. Redis IRedisTypedClient

    IRedisTypedClient IRedisTypedClient类相当于IRedicClient的强类型版,其方法与属性大多数与IRedisClient类似. 它支持在Redis中使用Linq查 ...

  9. ps查看进程

    ps:要对进程进行监测和控制,首先必须要了解当前进程的情况,也就是需要查看当前进程,而ps命令就是最基本同时也是非常强大的进程查看命令.使用该命令可以确定有哪些进程正在运行和运行的状态.进程是否结束. ...

  10. CSS多列布局Multi-column、伸缩布局Flexbox、网格布局Grid详解

    新css属性为我们提供了更加便捷的网页布局方式.来自微软的thomas lewis将带你认识去Grid Alignment,Flexibox Box以及Multi-column Layout这三大领域 ...