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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class IteratedLinearFunction {

	static long modPow(long a, long b, long MOD) {
		long sol = 1;
		while (b != 0) {
			if (b % 2 != 0) {
				sol = (sol * a) % MOD;
			}
			a = (a * a) % MOD;
			b /= 2;
		}
		return sol % MOD;
	}

	static long modInv(long a, long MOD) {
		return modPow(a, MOD - 2L, MOD);
	}

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(bf.readLine());
		long a = Long.parseLong(st.nextToken());
		long b = Long.parseLong(st.nextToken());
		long n = Long.parseLong(st.nextToken());
		long x = Long.parseLong(st.nextToken());
		long mod = 1000000007;
		long aton = modPow(a, n, mod);
		if (n == 0L)
			System.out.println(x);
		else if (n == 1L)
			System.out.println(((a * x) % mod + b) % mod);
		else if (a == 1L)
			System.out.println((x + (b * (n % mod)) % mod) % mod);
		else
			System.out.println((((aton * x) % mod)
					+ (b * (((((1 - aton) % mod + mod) % mod) * modInv((((1 - a) % mod + mod) % mod), mod) % mod) % mod)
							% mod) % mod) % mod);
	}
}
