#include<bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define rep(N) for (int i = 0; i < N; ++i)
#define Rep(a, b) for (int i = a; i < b; ++i)
#define For(i, N) for (int i = 0; i < N; ++i)
#define all(v) v.begin(), v.end()
#define rev(v) v.rbegin(), v.rend()
#define makei(N) int N; cin >> N;
#define makel(N) ll N; cin >> N;
#define makes(s) string s; cin >> s;
#define maked(d) double d; cin >> d;
#define makev(v, N) vi v(N); rep(N)cin >> v[i];
#define mod 1000000007
#define distance(x1,y1,x2,y2) sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));

using namespace std;
using ll = long long int;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vs = vector<string>;
using pii = pair<int, int>;
using pis = pair<int, string>;
using msi = map<string, int>;

template<typename T>
void say(T s) {
	cout << s << "\n";
}

template<typename T>
void say(vector<T> s) {
	auto itr = s.begin();
	cout << *(itr++);
	while (itr != s.end()) {
		cout << " " << *(itr++);
	}
	cout << "\n";
}

void yn(bool b) {
	if (b)say("yes");
	else say("no");
}

void Yn(bool b) {
	if (b)say("Yes");
	else say("No");
}

void YN(bool b) {
	if (b)say("YES");
	else say("NO");
}

template<typename T>
void maxi(T & a, T b) {
	a = max(a, b);
}

template<typename T>
void mini(T & a, T b) {
	a = min(a, b);
}

void exact_say(double x) {
	cout << setprecision(numeric_limits<double>::max_digits10) << x << endl;
}

template<class T> void getVector(vector<T> & v) {
	for (int i = 0; i < v.size(); ++i)
		cin >> v[i];
}

int main() {
	int h, w, a, b;
	cin >> h >> w >> a >> b;
	say(h * w - (h / a * a) * (w / b * b));
}

### Prompt

Develop a solution in cpp to the problem described below:

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
#include<bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define rep(N) for (int i = 0; i < N; ++i)
#define Rep(a, b) for (int i = a; i < b; ++i)
#define For(i, N) for (int i = 0; i < N; ++i)
#define all(v) v.begin(), v.end()
#define rev(v) v.rbegin(), v.rend()
#define makei(N) int N; cin >> N;
#define makel(N) ll N; cin >> N;
#define makes(s) string s; cin >> s;
#define maked(d) double d; cin >> d;
#define makev(v, N) vi v(N); rep(N)cin >> v[i];
#define mod 1000000007
#define distance(x1,y1,x2,y2) sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));

using namespace std;
using ll = long long int;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vs = vector<string>;
using pii = pair<int, int>;
using pis = pair<int, string>;
using msi = map<string, int>;

template<typename T>
void say(T s) {
	cout << s << "\n";
}

template<typename T>
void say(vector<T> s) {
	auto itr = s.begin();
	cout << *(itr++);
	while (itr != s.end()) {
		cout << " " << *(itr++);
	}
	cout << "\n";
}

void yn(bool b) {
	if (b)say("yes");
	else say("no");
}

void Yn(bool b) {
	if (b)say("Yes");
	else say("No");
}

void YN(bool b) {
	if (b)say("YES");
	else say("NO");
}

template<typename T>
void maxi(T & a, T b) {
	a = max(a, b);
}

template<typename T>
void mini(T & a, T b) {
	a = min(a, b);
}

void exact_say(double x) {
	cout << setprecision(numeric_limits<double>::max_digits10) << x << endl;
}

template<class T> void getVector(vector<T> & v) {
	for (int i = 0; i < v.size(); ++i)
		cin >> v[i];
}

int main() {
	int h, w, a, b;
	cin >> h >> w >> a >> b;
	say(h * w - (h / a * a) * (w / b * b));
}

```