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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;


public class D {

	static StringTokenizer st;
	static BufferedReader br;
	static PrintWriter pw;
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		long A = nextLong();
		long B = nextLong();
		long n = nextLong();
		long x = nextLong();
		int mod = (int) (1e9+7);
		long ans = BigInteger.valueOf(A).modPow(BigInteger.valueOf(n), BigInteger.valueOf(mod)).longValue();
		ans = ans * x % mod;
		if (A==1) {
			ans = (ans + n % mod * B) % mod;
		}
		else {
			long t = BigInteger.valueOf(A).modPow(BigInteger.valueOf(n), BigInteger.valueOf(mod)).longValue();
			t = (t-1+mod) % mod;
			t = t * B % mod;
			long inv = BigInteger.valueOf(A-1).modInverse(BigInteger.valueOf(mod)).longValue();
			t = t * inv % mod;
			ans = (ans + t) % mod;
		}
		System.out.println(ans);
		pw.close();
	}
	private static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
	private static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
	private static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	private static String next() throws IOException {
		while (st==null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
}
