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: Accepted)

#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i < n; ++i)
#define repe(i, n) for(int i=0; i <= n; ++i)
#define repr(i, n) for(int i=n-1; i > 0; --i)
#define all(x) (x).begin(), (x).end()
#define pb(x) push_back(x)
using namespace std;
template <class T> bool chmax(T &a, const T &b) {if (a<b) {a = b; return 1;} return 0;}
template <class T> bool chmin(T &a, const T &b) {if (b<a) {a = b; return 1;} return 0;}
template <class T> using V=vector<T>;
typedef long long ll;
const int INF = 1e9;
const ll MOD = 1000000007;
const ll MAX = 510000;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

double temp(int t, int h) {
    double s = t-h*0.006;
    return s;
}

int main() {
    string s;
    cin >> s;
    int cnt = 0;
    bool res = false;
    if (s[0] =='A') {
        res = true;
    }
    if(res==true) {
        for(int i=2; i<s.size()-1; ++i) {
            if (s[i] == 'C') {
                cnt++;
            }
        }
    }
    if (cnt==1) {
        res = true;
    } else {
        res = false;
    }
    if (res==true) {
        for(int i=1; i<s.size(); ++i) {
            if (s[i] == 'C' ) continue;
            if (s[i] < 'a' || s[i] > 'z') {
                res = false;
                break;
            }
        }
    }
    if (res == true) {
        cout << "AC" <<"\n";
    } else {
        cout << "WA" <<"\n";
    }
}