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 bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)


def input(): return sys.stdin.readline().rstrip("\r\n")
def isPrime(n) :
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True
def SieveOfEratosthenes(n): 
    prime=[]
    primes = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        if (primes[p] == True): 
            prime.append(p)
            for i in range(p * p, n+1, p): 
                primes[i] = False
        p += 1
    return prime
def primefactors(n):
    fac=[]
    while(n%2==0):
        fac.append(2)
        n=n//2
    for i in range(3,int(math.sqrt(n))+2):
        while(n%i==0):
            fac.append(i)
            n=n//i
    if n>1:
        fac.append(n)
    return fac
def factors(n):
    fac=set()
    fac.add(1)
    fac.add(n)
    for i in range(2,int(math.sqrt(n))+1):
        if n%i==0:
            fac.add(i)
            fac.add(n//i)
    return list(fac)
def modInverse(a, m):
    m0 = m
    y = 0
    x = 1
 
    if (m == 1):
        return 0
 
    while (a > 1):
 
       
        q = a // m
 
        t = m
 
       
        m = a % m
        a = t
        t = y
 
       
        y = x - q * y
        x = t
 
    if (x < 0):
        x = x + m0
 
    return x
 
#------------------------------------------------------code by AD18
a, b, n, x = map(int, input().split())
MOD = 1000000007
if a==1:
    print((1*x+b*(n))%MOD)
else:
    c1=(pow(a,n,MOD))%MOD
    c2=b*(c1-1)*modInverse(a-1,MOD)
    ans=(c1*x+c2)%MOD
    print((ans)%MOD)