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
'''Codeforces Problem 678D'''
from time import process_time
from sys import argv

def extended_gcd(aa, bb):
    '''Extended GCD algorithm'''
    lastremainder, remainder = abs(aa), abs(bb)
    x, lastx, y, lasty = 0, 1, 1, 0
    while remainder:
        lastremainder, (quotient, remainder) = remainder, divmod(lastremainder, remainder)
        x, lastx = lastx - quotient*x, x
        y, lasty = lasty - quotient*y, y
    return lastremainder, lastx * (-1 if aa < 0 else 1), lasty * (-1 if bb < 0 else 1)

def modinv(a, m):
    '''Inverse of a modulo m'''
    g, x, _ = extended_gcd(a, m)
    if g != 1:
        raise ValueError
    return x % m

def solve():
    '''Solve the problem'''
    A, B, n, x = map(int, input().split())
    m = 10 ** 9 + 7
    if A > 1:
        ans = pow(A, n, m) * x + (pow(A, n, m) - 1) * modinv(A - 1, m) * B
        print(ans % m)
    else:
        print((x + B * n) % m)

if len(argv) > 1:
    startTime = process_time()
    solve()
    print('Elapsed Time:', process_time() - startTime)
else:
    solve()
