задача на до с рукода

/*
⠸⣷⣦⠤⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣀⣠⣤⠀⠀⠀
⠀⠙⣿⡄⠈⠑⢄⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣀⠔⠊⠉⣿⡿⠁⠀⠀⠀
⠀⠀⠈⠣⡀⠀⠀⠑⢄⠀⠀⠀⠀⠀⠀⠀⠀⠀⡠⠊⠁⠀⠀⣰⠟⠀⠀⠀⣀⣀
⠀⠀⠀⠀⠈⠢⣄⠀⡈⠒⠊⠉⠁⠀⠈⠉⠑⠚⠀⠀⣀⠔⢊⣠⠤⠒⠊⠉⡜
⠀⠀⠀⠀⠀⠀⠀⡽⠁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠩⡔⠊⠁⠀⠀⠀⠀⠀ ⠀⠇
⠀⠀⠀⠀⠀⠀⠀⡇⢠⡤⢄⠀⠀⠀⠀⠀⡠⢤⣄⠀⡇⠀⠀⠀⠀⠀⠀⠀ ⢰⠀
⠀⠀⠀⠀⠀⠀⢀⠇⠹⠿⠟⠀⠀⠤⠀⠀⠻⠿⠟⠀⣇⠀⠀⡀⠠⠄⠒⠊⠁⠀
⠀⠀⠀⠀⠀⠀⢸⣿⣿⡆⠀⠰⠤⠖⠦⠴⠀⢀⣶⣿⣿⠀⠙⢄⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⢻⣿⠃⠀⠀⠀⠀⠀⠀⠀⠈⠿⡿⠛⢄⠀⠀⠱⣄⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀⢸⠈⠓⠦⠀⣀⣀⣀⠀⡠⠴⠊⠹⡞⣁⠤⠒⠉⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠀⣠⠃⠀⠀⠀⠀⡌⠉⠉⡤⠀⠀⠀⠀⢻⠿⠆⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠰⠁⡀⠀⠀⠀⠀⢸⠀⢰⠃⠀⠀⠀⢠⠀⢣⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⠀⢶⣗⠧⡀⢳⠀⠀⠀⠀⢸⣀⣸⠀⠀⠀⢀⡜⠀⣸⢤⣶⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠈⠻⣿⣦⣈⣧⡀⠀⠀⢸⣿⣿⠀⠀⢀⣼⡀⣨⣿⡿⠁⠀⠀⠀⠀⠀⠀
⠀⠀⠀⠀⠀⠈⠻⠿⠿⠓⠄⠤⠘⠉⠙⠤⢀⠾⠿⣿⠟⠋
*/


#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <stdio.h>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <iterator>
#include <map>
#include <iomanip>
#include <unordered_set>
using namespace std;

const unsigned int R = 1 << 15, INF = 4294967294;
//unsigned int MOD = 4294967296;

unsigned int tree[2 * R + 5];

inline void build()
{
    for (int i = R - 1; i > 0; i--)
    {
        tree[i] = min(tree[i * 2], tree[i * 2 + 1]);
    }

    return;
}

inline unsigned int minn(unsigned int ql, unsigned int qr, unsigned int v, unsigned int nl, unsigned int nr)
{
    if (nl > qr || nr < ql)
    {
        return INF;
    }

    if (nl >= ql && nr <= qr)
    {
        return tree[v];
    }

    unsigned int nm = (nl + nr) / 2;

    unsigned int lol = minn(ql, qr, 2 * v, nl, nm);

    unsigned int kek = INF;

    if (tree[v * 2 + 1] < lol || lol == INF)
    {
        kek = minn(ql, qr, 2 * v + 1, nm + 1, nr);
    }

    return min(lol, kek);
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    unsigned int n, m, a, b;

    while (true)
    {
        cin >> n >> m >> a >> b;

        if (n == m && m == a && a == b && b == 0)
        {
            return 0;
        }

        for (unsigned int i = 0; i < n; i++)
        {
            tree[i + R] = (a * (i + 1) + b);
        }

        for (unsigned int i = n; i <= R; i++)
        {
            tree[i + R] = INF;
        }

        build();

        unsigned long long ans = 0;

        for (unsigned int i = n + 1; i < n + 2 * m; i += 2)
        {
            unsigned int l = (a * i + b);
            unsigned int r = (a * (i + 1) + b);

            l %= n;
            r %= n;

            l++;
            r++;

            ans += minn(min(l, r), max(l, r), 1, 1, R);
        }

        cout << ans << "\n";
    }
}