Problem:
Score : 200 points

Problem Statement

You are given a string S. Each character of S is uppercase or lowercase English letter.
Determine if S satisfies all of the following conditions:

The initial character of S is an uppercase A.

There is exactly one occurrence of C between the third character from the beginning and the second to last character (inclusive).

All letters except the A and C mentioned above are lowercase.

Constraints

4 ≤ |S| ≤ 10 (|S| is the length of the string S.)

Each character of S is uppercase or lowercase English letter.

Input

Input is given from Standard Input in the following format:

S

Output

If S satisfies all of the conditions in the problem statement, print AC; otherwise, print WA.

Sample Input 1

AtCoder

Sample Output 1

AC

The first letter is A, the third letter is C and the remaining letters are all lowercase, so all the conditions are satisfied.

Sample Input 2

ACoder

Sample Output 2

WA

The second letter should not be C.

Sample Input 3

AcycliC

Sample Output 3

WA

The last letter should not be C, either.

Sample Input 4

AtCoCo

Sample Output 4

WA

There should not be two or more occurrences of C.

Sample Input 5

Atcoder

Sample Output 5

WA

The number of C should not be zero, either.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)


#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define endl "\n"
#define pb push_back
#define f(i,n) for(i=0;i<n;i++)
#define F(i,a,b) for(i=a;a<=b;i++)
#define arr(a,n) for( i=0;i<n;i++)cin>>a[i];
#define fi first
#define se second
#define mp make_pair
#define mod 1000000007
#define YES cout<<"YES"<<endl;
#define Yes cout<<"Yes"<<endl;
#define NO cout<<"NO"<<endl;
#define No cout<<"No"<<endl;
#define yes cout<<"yes"<<endl;
#define no cout<<"no"<<endl;
#define vi vector<ll>
#define ed end()
#define bg begin()
#define sz size()
#define ln length()
#define s() sort(a,a+n);
#define sr() sort(a,a+n,greater<ll>());
#define v()  sort(v.begin(),v.end());
#define vr() sort(v.begin(),v.end(),greater<ll>());
#define mod 1000000007
#define fast() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
ll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);}
ll power(ll x,ll y,ll p){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}
int main() {
   /* #ifndef ONLINE_JUDGE
        // for getting input from input.txt
        freopen("input.txt", "r", stdin);
        // for writing output to output.txt
        freopen("output.txt", "w", stdout);
    #endif*/
        /* 
        Ofcourse it's Hard.
        It's supposed to be Hard.
        If it's easy everyone would do it.
        HARD IS WHAT MAKES IT GREAT

        YESTERDAY U SAID TOMORROW

        SLOWLY BECOMING THE PERSON I 
        SHOULD HAVE BEEN A LONG TIME AGO

        SAME TASK CAN'T BE FOUND DIFFICULT TWICE

        BTBHWSITW

        SPRH TU KAB P

        CP IS ALL ABOUT THINKING 
        YOU HAVE MUCH MORE POTENTIAL THAN U THINK

        AJIT SHIDDAT 10

        UR DAILY ROUTINE 


        */

fast();
//ll t;cin>>t;while(t--)
{
  string s;
  cin>>s;
  ll n=s.length(),f=0,cou=0;
  if(s[0]!='A')
    f=1;
  for(int i=2;i<n-1;i++)
  {
    if(s[i]=='C')
      cou++;
    if(s[i]<97&&s[i]!='C')
      f=1;
    if(cou>1)
      f=1;
  }
  if(f)
    cout<<"WA"<<endl;
  else
    cout<<"AC"<<endl;

    
}
return 0;
}