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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class D {

	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);
	}

	static long fix(long n, long mod) {
		return (n % mod + mod) % mod;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pr = new PrintWriter(System.out);
		long A = sc.nextLong();
		long B = sc.nextLong();
		long n = sc.nextLong();
		long x = sc.nextLong();
		long MOD = 1000000007;
		long answer = 0;
		long st = 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 {
			answer = (st * x) % MOD;
			long f = fix(1 - st, MOD);
			long v = modInv(fix(1 - A, MOD), MOD);
			long r = ((f % MOD) * (v % MOD)) % MOD;
			long k = B * r;
			System.out.println((answer % MOD + k % MOD) % MOD);
		}
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}
