//実装が重そうな問題はある程度考えてから書く
//初期化を忘れずに（特に二分探索とか）
//コーナーケースを考えて（特に場合分けとか）
//不可解すぎるバグは配列外参照（配列の長さ）を検討
#include "bits/stdc++.h"
#define YES "YES"
#define NO "NO"
#define YESNO OUT(three(solve(),YES,NO))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(a);i>=(LL)(b);i--)
#define REP(i,b) FOR(i,zero,b)
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,b-1,zero)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB emplace_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define int long long
#define zero 0LL
#define pb PB
using namespace std;
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef priority_queue<pr, Vpr ,greater<pr> > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
int dx[] = { 0,1,0,-1,0 }, dy[] = { 1,0,-1,0,0 };
const int SIZE1 =3e5 + 1000;
const int SIZE2 = 5010;
const int SIZE3 = 22;
const int SIZE = SIZE1;
const LL p = 7 + 1e9;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
ut N, M, K, X, L, Y,H,W,Q;
// ut A,B,C,D,E,F,G,H,I,J,O,P,Q,R,T,U;
VI edges[SIZE];
LL vals[SIZE], nums[SIZE], maps[SIZE3][SIZE3], answer = zero;
LL A[SIZE], B[SIZE];
LL solve() {
	LL a, b;
	cin >> H >> W >> a >> b;
	cout << (H%a)*W + H * (W%b) - (H%a)*(W%b) << endl;
	return 0;
}
signed main() {
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(10);

	solve();
	cin >> N;
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
//実装が重そうな問題はある程度考えてから書く
//初期化を忘れずに（特に二分探索とか）
//コーナーケースを考えて（特に場合分けとか）
//不可解すぎるバグは配列外参照（配列の長さ）を検討
#include "bits/stdc++.h"
#define YES "YES"
#define NO "NO"
#define YESNO OUT(three(solve(),YES,NO))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(a);i>=(LL)(b);i--)
#define REP(i,b) FOR(i,zero,b)
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,b-1,zero)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB emplace_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define int long long
#define zero 0LL
#define pb PB
using namespace std;
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef long long LL;
typedef double ld;
typedef LL ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef priority_queue<pr, Vpr ,greater<pr> > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
int dx[] = { 0,1,0,-1,0 }, dy[] = { 1,0,-1,0,0 };
const int SIZE1 =3e5 + 1000;
const int SIZE2 = 5010;
const int SIZE3 = 22;
const int SIZE = SIZE1;
const LL p = 7 + 1e9;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
ut N, M, K, X, L, Y,H,W,Q;
// ut A,B,C,D,E,F,G,H,I,J,O,P,Q,R,T,U;
VI edges[SIZE];
LL vals[SIZE], nums[SIZE], maps[SIZE3][SIZE3], answer = zero;
LL A[SIZE], B[SIZE];
LL solve() {
	LL a, b;
	cin >> H >> W >> a >> b;
	cout << (H%a)*W + H * (W%b) - (H%a)*(W%b) << endl;
	return 0;
}
signed main() {
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(10);

	solve();
	cin >> N;
	return 0;
}

```