problem

I want to put as many rectangular tiles as possible on a rectangular wall with a size of $ h $ in height and $ w $ in width, and a size of $ a $ in height and $ b $ in width.

The following conditions must be met when attaching tiles.


* Do not stack tiles.
* Do not apply tiles diagonally, that is, any edge of the tile is parallel or perpendicular to any edge of the wall.
* Do not change the orientation of the tiles, that is, do not swap the vertical and horizontal directions.



When as many tiles as possible are pasted, find the sum of the areas not covered by the tiles.



output

Output the total area of ​​the part not covered by the tile. Also, output a line break at the end.

Example

Input

5 8
2 2


Output

8
#include <bits/stdc++.h>
using namespace std;
//#include <boost/multiprecision/cpp_int.hpp>
//using multiInt = boost::multiprecision::cpp_int;

using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<typename Q_temp>
using smaller_queue = priority_queue <Q_temp, vector<Q_temp>, greater<Q_temp> >;

const int INF = (int) 1e9;
const ll LINF = (ll) 1e18;
const ll MOD = (ll) (1e9 + 7);
const double PI = acos(-1.0);
const int limit = 200010;

#define REP(i,m,n) for(ll i = m; i < (ll)(n); ++i)
#define rep(i,n) REP(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << endl
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbg2(x_) for(auto a_ : x_) cerr << a_ << " "; cerr << endl;
#define dbg3(x_ , sx_) rep(i, sx_) cerr << x_[i] << " "; cerr << endl;
vector<int> Dx = {0, 0, -1, 1, -1, 1, -1, 1, 0};
vector<int> Dy = {1, -1, 0, 0, -1, -1, 1, 1, 0};

//------------------------------------------------------

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int h, w, a, b;
    cin >> h >> w >> a >> b;
    cout << h * w - (h - h % a) * (w - w % b) << endl;
    return 0;
}
