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
#http://codeforces.com/problemset/problem/678/D
def matrixProductModulo(dic1,dic2,modulo):
    dic = {}
    dic['a'] = (dic1['a']*dic2['a'] + dic1['b']*dic2['c'])%modulo
    dic['b'] = (dic1['a']*dic2['b'] + dic1['b']*dic2['d'])%modulo
    dic['c'] = (dic1['c']*dic2['a'] + dic1['d']*dic2['c'])%modulo
    dic['d'] = (dic1['c']*dic2['b'] + dic1['d']*dic2['d'])%modulo
    return dic
def getIdentityMatrix():
    identityMatrix = {}
    identityMatrix['a'] = 1
    identityMatrix['b'] = 0
    identityMatrix['c'] = 0
    identityMatrix['d'] = 1
    return identityMatrix
    
def binaryExp(dic,power,modulo):
    if power == 0:
        return getIdentityMatrix()
    halfPower = binaryExp(dic,power/2,modulo)
    halfPowerProduct = matrixProductModulo(halfPower,halfPower,modulo)
    if power%2 == 0:
        return halfPowerProduct
    else:
        return matrixProductModulo(dic,halfPowerProduct,modulo)
        
    
a,b,n,x = map(int,raw_input().split())
modulo = 10**9 + 7
matrix = {}
matrix['a'] = a
matrix['b'] = b
matrix['c'] = 0
matrix['d'] = 1
matrixPower = binaryExp(matrix,n,modulo)
ans = (matrixPower['a']*x + matrixPower['b'])%modulo
print ans
