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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
#include <bitset>
#define M 1000000007
#define M1 998244353

#define tr(container, it) \
for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

#define fast                        \
  ios_base::sync_with_stdio(false); \
  cin.tie(NULL);

#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

ll power(ll x, ll n)
{
    ll result = 1;
    while (n)
    {
        if (n % 2 == 1)
            result = result * x;
        n = n / 2;
        x = x * x;
    }
    return result;
}

ll gcd(ll a, ll b)
{
    if (!b)
        return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return (a * b) / gcd(a, b);
}

inline ll mul(ll x, ll y, ll m)
{
    ll z = 1LL * x * y;
    if (z >= m)
    {
        z %= m;
    }
    return z;
}

ll powmod(ll x, ll y, ll m)
{
    ll r = 1;
    while (y)
    {
        if (y & 1)
        {
            r = mul(r, x, m);
        }
        y >>= 1;
        x = mul(x, x, m);
    }
    return r;
}



void start()
{
    fast;
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif
}
/***********************************************************************************************************/

int main()
{
    start();
    string s;
    cin>>s;
    int n=s.length();
    if (s[0]!='A')
    {
        cout<<"WA";
        return 0;
    }
    int c=0;
    for (int i=1;i<n;i++)
    {
        if (s[i]=='C')
        {
            c++;
            if (i<=1||i>=n-3||c>1)
            {
                cout<<"WA";
                return 0;
            }
        }
    }
    if (c==1)
    {
        cout<<"AC";
    }
    else
    {
        cout<<"WA";
    }

    return 0;
}