Problem:
Score : 100 points

Problem Statement

We have A balls with the string S written on each of them and B balls with the string T written on each of them.

From these balls, Takahashi chooses one with the string U written on it and throws it away.

Find the number of balls with the string S and balls with the string T that we have now.

Constraints

S, T, and U are strings consisting of lowercase English letters.

The lengths of S and T are each between 1 and 10 (inclusive).

S \not= T

S=U or T=U.

1 \leq A,B \leq 10

A and B are integers.

Input

Input is given from Standard Input in the following format:

S T
A B
U

Output

Print the answer, with space in between.

Sample Input 1

red blue
3 4
red

Sample Output 1

2 4

Takahashi chose a ball with red written on it and threw it away.
Now we have two balls with the string S and four balls with the string T.

Sample Input 2

red blue
5 5
blue

Sample Output 2

5 4

Takahashi chose a ball with blue written on it and threw it away.
Now we have five balls with the string S and four balls with the string T.

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

#include <bits/stdc++.h>
#include <string>
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <bitset>
typedef long long ll;
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
 
const int mod = 1000000007;
//const int mod = 998244353;
 

struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
  mint operator+(const mint a) const { return mint(*this) += a;}
  mint operator-(const mint a) const { return mint(*this) -= a;}
  mint operator*(const mint a) const { return mint(*this) *= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const { return pow(mod-2);}
  mint& operator/=(const mint a) { return *this *= a.inv();}
  mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
 
 
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
} c(200005);
 
vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}
vector<long long> enum_divisors(long long N) {
    vector<long long> res;
    for (long long i = 1; i * i <= N; ++i) {
        if (N % i == 0) {
            res.push_back(i);
            // 重複しないならば i の相方である N/i も push
            if (N/i != i) res.push_back(N/i);
        }
    }
    // 小さい順に並び替える
    sort(res.begin(), res.end());
    return res;
}

// M * (M-1)^(N-1-K) * N-1 C K
int main(int argc, const char * argv[]) {
    string S,T;
    cin >> S >> T;
    ll A,B; cin >> A >> B;
    string U; cin >> U;
    
    if(U.compare(S)==0){
        cout << A -1 << " " << B << endl;
    }else{
        cout << A << " " << B-1<< endl;
    }
    return 0;
}