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
def calcInverse(r2, r1):
    global Mod
    s1, s2, t1, t2 = 1, 0, 0, 1

    while r2 != 0:
        q = r1 / r2
        r = r1 % r2
        s = s1 - q * s2
        t = t1 - q * t2

        r1, r2 = r2, r
        s1, s2 = s2, s
        t1, t2 = t2, t
    return t1 % Mod



def calcExponent(a, n):
    global Mod

    n = bin(n)[2:]
    Len = len(n) - 1
    tmp = a

    Sum = 1
    while Len >= 0:
        if n[Len] == '1': Sum = (Sum * tmp) % Mod
        tmp = (tmp * tmp) % Mod
        Len -= 1
    return Sum % Mod
    
    
    
## Entry of Program
#
#

Mod = 10 ** 9 + 7
a, b, n, x = [int(x) for x in raw_input().split(' ')]

if a == 1: print (n * b + x) % Mod
else: print ((b * (calcExponent(a, n) - 1) * calcInverse(a - 1, Mod)) % Mod + calcExponent(a, n) * x) % Mod
