#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

constexpr int N = 105;
constexpr int inf = 1e9;

int f[N][N][2];

void solve() {
    int x, y, p, q;
    cin >> x >> y >> p >> q;

    for (int i = 0; i <= x; i++) {
        for (int j = 0; j <= y; j++) {
            f[i][j][0] = f[i][j][1] = -1;
        }
    }

    //f[i][j][0/1]表示i只羊 j只狼 现在在岸的某一边的
    f[x][y][0] = 0;
    queue<array<int, 3>> que;
    que.push({x, y, 0});

    while (!que.empty()) {
        auto [nx, ny, k] = que.front();
        que.pop();

        //还有nx个羊 ny只狼 0/1代表左右侧

        if (k == 0) {
            //枚举运走i只羊 j只狼
            for (int i = 0; i <= p; i++) {
                for (int j = 0; i + j <= p; j++) {
                    if (i > nx || j > ny) {
                        continue;
                    }

                    int nnx = nx - i, nny = ny - j; //留下nnx只羊 nny只狼
                    if (nnx != 0 && nny > nnx + q) {
                        continue;
                    }

                    if (f[nnx][nny][1] == -1) {
                        f[nnx][nny][1] = f[nx][ny][0] + 1;
                        que.push({nnx, nny, 1});
                    }
                }
            }
        } else {
            for (int i = 0; i <= p; i++) {
                for (int j = 0; i + j <= p; j++) {
                    if (i > x - nx || j > y - ny) {
                        continue;
                    }


                    int nnx = x - nx - i, nny = y - ny - j;
                    if (nnx != 0 && nny > nnx + q) {
                        continue;
                    }


                    if (f[nx + i][ny + j][0] == -1) {
                        f[nx + i][ny + j][0] = f[nx][ny][1] + 1;
                        que.push({nx + i, ny + j, 0});
                    }
                }
            }

        }

    }

    constexpr int inf = 1e9;
    int ans = inf;
    for (int j = 0; j <= y; j++) {
        if (f[0][j][0] != -1) {
            ans = min(ans, f[0][j][1]);
        }
    }

    if (ans == inf) {
        ans = -1;
    }

    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int T = 1;
    // cin >> T;
    
    while (T--) {
        solve();
    }
    return 0;
}