Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values A, B, n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers A, B, n and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples

Input

3 4 1 1


Output

7


Input

3 4 2 1


Output

25


Input

3 4 3 1


Output

79
import java.util.Scanner;

public class D
{
	static long MOD = (long)1e9 + 7;
	public static long[][] mult(long[][] a, long[][] b)
	{
		int R = a.length, C = b[0].length;
		long[][] ret = new long[R][C];
		for (int r = 0; r < R; ++r) for (int c = 0; c < C; ++c) 
			for (int i = 0; i < a[0].length; ++i) ret[r][c] = (ret[r][c] + (a[r][i]*b[i][c])%MOD)%MOD;
		return ret;
	}
	public static long[][] eye(int n)
	{
		long[][] ret = new long[n][n];
		for (int i = 0; i < ret.length; ++i) ret[i][i] = 1;
		return ret;
	}
	public static long[][] pow(long[][] a, long p)
	{
		if (p == 0) return eye(a.length);
		if (p % 2 == 1) return mult(pow(a,p-1),a);
		long[][] sqrt = pow(a,p/2);
		return mult(sqrt,sqrt);
	}
	static long f(long A, long B, long x) { return ((A*x)%MOD + B)%MOD; } 
	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		long A = in.nextLong();
		long B = in.nextLong();
		long n = in.nextLong();
		long x = in.nextLong();
		long[][] init = {{x},{1}};
		long[][] mat = {{A,B},{0,1}};
		long[][] fin = mult(pow(mat,n),init);
		System.out.println(fin[0][0]);
	}
}
