using System;
using static System.Console;
using System.Linq;
using System.Collections.Generic;

class Program
{
    static int NN => int.Parse(ReadLine());
    static long[] NList => ReadLine().Split().Select(long.Parse).ToArray();
    static long[] LList(long n) => Enumerable.Repeat(0, (int)n).Select(_ => long.Parse(ReadLine())).ToArray();
    public static void Main()
    {
        Solve();
    }
    static void Solve()
    {
        var c = NList;
        var (n, m, k, p, q) = (c[0], c[1], c[2], c[3], c[4]);
        var b = LList(n);

        var a = new long[][] { new long[] { 0, 0 }};
        for (var i = 0; i < m; ++i) a[0][0] += b[i];
        for (var i = m; i < n; ++i) a[0][1] += b[i];
        a[0][0] %= mod;
        a[0][1] %= mod;
        var revq = Exp(q, mod - 2, mod);
        var x = new long[][] { new long[] { (q - p) * revq % mod, p * revq % mod }, new long[] { p * revq % mod, (q - p) * revq % mod } };
        WriteLine(Mul(a, Pow(x, k))[0][0]);
    }
    static int mod = 1_000_000_007;
    // 行列の累乗
    public static long[][] Pow(long[][] m, long k)
    {
        var multi = m;
        var r = new long[m.Length][];
        for (var i = 0; i < m.Length; ++i)
        {
            r[i] = new long[m.Length];
            r[i][i] = 1;
        }
        while (k > 0)
        {
            if ((k & 1) == 1) r = Mul(r, multi);
            multi = Mul(multi, multi);
            k >>= 1;
        }
        return r;
    }
    // 行列の積
    public static long[][] Mul(long[][] x, long[][] y)
    {
        var r = new long[x.Length][];
        for (var i = 0; i < x.Length; ++i) r[i] = new long[y[0].Length];
        for (var i = 0; i < x.Length; ++i) for (var k = 0; k < x[0].Length; ++k)
        {
            for (var j = 0; j < y[0].Length; ++j) r[i][j] = (r[i][j] + x[i][k] * y[k][j]) % mod;
        }
        return r;
    }
    static long Exp(long n, long p, int mod)
    {
        long _n = n % mod;
        var _p = p;
        var result = 1L;
        if ((_p & 1) == 1) result *= _n;
        while (_p > 0)
        {
            _n = _n * _n % mod;
            _p >>= 1;
            if ((_p & 1) == 1) result = result * _n % mod;
        }
        return result;
    }
}