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. 【leetcode刷题笔记】Flatten Binary Tree to Linked List

    Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 T ...

  2. POJ2406Power Strings (最小循环节)(KMP||后缀数组)

    Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc&quo ...

  3. 洛谷【P1080】国王游戏

    我对贪心的理解:https://www.cnblogs.com/AKMer/p/9776293.html 题目传送门:https://www.luogu.org/problemnew/show/P10 ...

  4. MySQL Explain详解(转)

    explain SELECT a.* FROM test a,(select id from test where level_id <=4 order by aa_id limit 24300 ...

  5. NorthSJ项目零碎知识点

    1 div显示纵向滚动条: <div style="overflow-y:auto;overflow-x:hidden;"></div> 2 FineUI的 ...

  6. POJ3087(模拟)

    #include"iostream" #include"string" #include"map" using namespace std; ...

  7. hashCode之二--Java:重写equals()和hashCode()

    以下内容总结自<Effective Java>. 1.何时需要重写equals() 当一个类有自己特有的“逻辑相等”概念(不同于对象身份的概念). 2.设计equals() [1]使用in ...

  8. 微服务理论之三:RPC框架原理

    RPC调用是面向服务架构场景下进行服务间调用的常用组件,一个完整的RPC调用的流程如图1所示: 图1 RPC调用流程 为了方便RPC调用者和服务者的开发,开发者们开发了很多RPC框架.比较有名的RPC ...

  9. js字符串API

    1.charAt(n) :返回字符串n位置的字符 2.substr(n,m):n:开始截取的位置 m:截取的长度 2.substring(n,m):n:开始截取的位置 m:截取结束的位置 3.repl ...

  10. 微服务学习一 微服务session 管理

    集群和分布式架构中: session管理有三种方法: 1: Cookie: 将Session对象保存在Cookie,保存在浏览器端.浏览器发送请求的时候,会把整个session放在请求里一起发送到se ...