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 <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm> 
using namespace std;

#define ls st<<1
#define rs st<<1|1
#define LL long long 
#define CLR(a,b) memset(a,(b),sizeof(a))

const int MAXN = 1e3+11;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f; 

/* ------------------------基本函数-------------------------*/
long long read()
{
    long long res = 0;
    char ch = 0;
    while (ch < '0' || ch > '9')
        ch = getchar();
    while (ch >= '0' && ch <= '9')
    {
        res = res * 10 + ch - '0';
        ch = getchar();
    }
    return res;
}
bool prime(int x) {
	if(x==2 || x==3) 
		return true;
	if(x%2==0 || x%3==0 || x==1)
		return false;
	int k = sqrt(x)+1;
	for(int i = 5; i < k; i++) {
		if(x%i==0 || x%(i+2)==0)
			return false;
	}
	return true;
} 

LL pow_mod(int a, int b, int c) {
	LL ans = 1;
	while(b) {
		if(b&1)
			ans = (ans*1) % c;
		a = (a*a) % c;
		b /= 2;
	}
	return ans;
}
int gcd(int a, int b) {
	if(!b) 
		return a;
	return gcd(b,a%b);
}
/*-------------------------------------------------------*/


int main() {
	LL x,y,a,b;
	cin >> x >> y >> a >> b;
	cout << max(x*y,a*b) << endl;
return 0;
}