Problem:
Score : 100 points

Problem Statement

There are two rectangles.
The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B.
The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D.

Print the area of the rectangle with the larger area.
If the two rectangles have equal areas, print that area.

Constraints

All input values are integers.

1≤A≤10^4

1≤B≤10^4

1≤C≤10^4

1≤D≤10^4

Input

The input is given from Standard Input in the following format:

A B C D

Output

Print the area of the rectangle with the larger area.
If the two rectangles have equal areas, print that area.

Sample Input 1

3 5 2 7

Sample Output 1

15

The first rectangle has an area of 3×5=15, and the second rectangle has an area of 2×7=14.
Thus, the output should be 15, the larger area.

Sample Input 2

100 600 200 300

Sample Output 2

60000

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

#include <bits/stdc++.h>
#define pb push_back
#define ll long long
#define mem(x, val) memset(x, val, sizeof(x))
#define mk make_pair
#define f(i,n) for(i = 0; i < n; i++)
#define f1(i,n) for(i = 1; i <= n; i++)
#define all(a) a.begin(),a.end()
#define upp(v,val) upper_bound(v.begin(), v.end(), val)
#define lower(v,val) lower_bound(v.begin(), v.end(), val)
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define dbg4(x,y,z,q) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << ", " << #q << " = " << q << endl
#define jaldi_chal() ios_base::sync_with_stdio(false); cin.tie(NULL);
#define S second
#define F first
#define fm(it,m) for(it = m.begin(); it != m.end(); it++)
#define ct_set(n) __builtin_popcount(n) // count no of set bits
#define INF 0x3f3f3f3f
#define endl '\n'
#define mod 1000000007
#define PI 3.14159265    // acosl(-1)
#define precision cout << setprecision(15)
#define print_arr(a,n) for(int i = 0; i < n; i++) cout << a[i] << " ";
#define print_vec(v) for(int i = 0; i < v.size(); i++) cout << v[i] << " ";
//vec.erase(std::remove(vec.begin(), vec.end(), val), vec.end());
using namespace std;

// v.resize(unique(v.begin(), v.end()) - v.begin());   // for unique elements in sorted vector
// fflush(stdout);  // after every output

// Using 128 bit integers
/*
	std::string toString(__int128 num) {
    std::string str;
    do {
        int digit = num % 10;
        str = std::to_string(digit) + str;
	        num = (num - digit) / 10;
	    } while (num != 0);
	    return str;
	}
	
	int main() {
	    __int128_t x;
	    string s;
	    cin >> s;
	    x = (stoll)(s);
	    __int128_t ans = x * x * x;
	    cout << toString(ans);
	}

*/

bool isPrime(ll n) {
    if (n <= 1)
        return false;
    if (n <= 3)
        return true;
 
    if (n % 2 == 0 || n % 3 == 0)
        return false;
 
    for (ll i = 5; i * i <= n; i = i + 6)
        if (n % i == 0 || n % (i + 2) == 0)
            return false;
 
    return true;
}

ll gcd(ll a,ll b) {
	if (a == 0)    
		return b; 	
	return gcd(b % a, a); 
}

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

ll add(ll a, ll b) {
	a += b;
	if (a >= mod) a -= mod;
	return a;
}
 
ll sub(ll a, ll b) {
	a -= b;
	if (a < 0) {
		a += mod;
	}
	return a;
}
 
ll mul(ll a, ll b) {
	return (a * b) % mod;
}

vector <ll> primes;

bool prime[10005];
void seive() {
	mem(prime, 1);
	prime[0] = 0;
	prime[1] = 0;
	for(ll i = 2; i <= 10000; i++)
	{
		if(prime[i] == 1)
		{
			for(ll j = i * i; j <= 10000; j += i)
			prime[j] = 0;
		}
	}	
}

ll power(ll a, ll b) {
	ll ans = 1;
	while(b > 0)
	{
		if(b % 2 == 1)
		ans = (ans % mod * a % mod) % mod;
		a = (a * a) % mod;
		b = b / 2;
	}
	return ans;
}

template <typename T>
std::string NumberToString ( T Number ) {
	std::ostringstream ss;
	ss << Number;
	return ss.str();
}

string num_to_bits(ll n) {
	string ans = "";
	while(n) {
		ans += (n % 2) + '0';
		n /= 2;
	}
	reverse(all(ans));
	return ans;
}

ll bits_to_num(string s) {
	ll ans = 0;
	for(ll i = 0; i < s.size(); i++) {
		ll x = s.size() - i - 1;
		ll val = pow(2LL, x);
		if(s[i] == '1') {
			ans = ans + val;
		}
	}
	return ans;
}

ll gcdExtended(ll a, ll b, ll *x, ll *y);   
ll modInverse(ll b, ll m) { 
    ll x, y; 
    ll g = gcdExtended(b, m, &x, &y); 
    if (g != 1) 
        return -1; 
    return (x % m + m) % m; 
} 
  
// Function to compute a/b under modlo m 
ll modDivide(ll a, ll b, ll m) { 
    a = a % m; 
    ll inv = modInverse(b, m); 
    //if (inv == -1) 
       //cout << "Division not defined"; 
    //else
       //cout << "Result of division is " << (inv * a) % m; 
       return (inv * a) % m;
} 

ll gcdExtended(ll a, ll b, ll *x, ll *y) { 
    if (a == 0) 
    { 
        *x = 0, *y = 1; 
        return b; 
    } 
    ll x1, y1; 
    ll gcd = gcdExtended(b % a, a, &x1, &y1); 
    *x = y1 - (b / a) * x1; 
    *y = x1; 
    return gcd; 
} 
// To call for modulo division (a / b) % m : modDivide(a, b, m);

/* This will store divisors of the number from 1 to n
	for(i = 1; i <= n; i++) {
		for(j = i; j <= n; j += i) {
			v[j].pb(i);
		}
	}
*/

void solve() {
	
	ll a, b, c, d;
	cin >> a >> b >> c >> d;
	cout << max(a * b, c * d);
	
}

int main() {
	jaldi_chal()
	
//	freopen("input.txt", "r" , stdin);
//  freopen("output.txt", "w", stdout);
	
//	seive();
//	for(int i = 2; i <= 10000; i++)	{
//		if(prime[i])
//		primes.pb(i);
//	}
	
	bool codechef = 0;
	ll t = 1;
	
	if(codechef) {
		cin >> t;
	}
	
	while(t--) {
		solve();
	}
	
	return 0;
}