[THE SxPLAY & KIVΛ - 漂流](https://soundcloud.com/kivawu/hyouryu)

[KIVΛ & Nikki Simmons - Perspectives](https://soundcloud.com/kivawu/perspectives)

With a new body, our idol Aroma White (or should we call her Kaori Minamiya?) begins to uncover her lost past through the OS space.

The space can be considered a 2D plane, with an infinite number of data nodes, indexed from 0, with their coordinates defined as follows:

  * The coordinates of the 0-th node is (x_0, y_0) 
  * For i > 0, the coordinates of i-th node is (a_x ⋅ x_{i-1} + b_x, a_y ⋅ y_{i-1} + b_y) 



Initially Aroma stands at the point (x_s, y_s). She can stay in OS space for at most t seconds, because after this time she has to warp back to the real world. She doesn't need to return to the entry point (x_s, y_s) to warp home.

While within the OS space, Aroma can do the following actions:

  * From the point (x, y), Aroma can move to one of the following points: (x-1, y), (x+1, y), (x, y-1) or (x, y+1). This action requires 1 second. 
  * If there is a data node at where Aroma is staying, she can collect it. We can assume this action costs 0 seconds. Of course, each data node can be collected at most once. 



Aroma wants to collect as many data as possible before warping back. Can you help her in calculating the maximum number of data nodes she could collect within t seconds?

Input

The first line contains integers x_0, y_0, a_x, a_y, b_x, b_y (1 ≤ x_0, y_0 ≤ 10^{16}, 2 ≤ a_x, a_y ≤ 100, 0 ≤ b_x, b_y ≤ 10^{16}), which define the coordinates of the data nodes.

The second line contains integers x_s, y_s, t (1 ≤ x_s, y_s, t ≤ 10^{16}) – the initial Aroma's coordinates and the amount of time available.

Output

Print a single integer — the maximum number of data nodes Aroma can collect within t seconds.

Examples

Input


1 1 2 3 1 0
2 4 20


Output


3

Input


1 1 2 3 1 0
15 27 26


Output


2

Input


1 1 2 3 1 0
2 2 1


Output


0

Note

In all three examples, the coordinates of the first 5 data nodes are (1, 1), (3, 3), (7, 9), (15, 27) and (31, 81) (remember that nodes are numbered from 0).

In the first example, the optimal route to collect 3 nodes is as follows: 

  * Go to the coordinates (3, 3) and collect the 1-st node. This takes |3 - 2| + |3 - 4| = 2 seconds. 
  * Go to the coordinates (1, 1) and collect the 0-th node. This takes |1 - 3| + |1 - 3| = 4 seconds. 
  * Go to the coordinates (7, 9) and collect the 2-nd node. This takes |7 - 1| + |9 - 1| = 14 seconds. 



In the second example, the optimal route to collect 2 nodes is as follows: 

  * Collect the 3-rd node. This requires no seconds. 
  * Go to the coordinates (7, 9) and collect the 2-th node. This takes |15 - 7| + |27 - 9| = 26 seconds. 



In the third example, Aroma can't collect any nodes. She should have taken proper rest instead of rushing into the OS space like that.
import java.io.*;
import java.util.*;
import java.math.*;
import java.awt.Point;
 
public class Code {
	//static final long MOD = 998244353L;
	//static final long INF = -1000000000000000007L;
	static final long MOD = 1000000007L;
	//static final int INF = 1000000007;

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		long[][] points = new long[65][2];
		points[0][0] = sc.nl();
		points[0][1] = sc.nl();
		long Ax = sc.nl();
		long Ay = sc.nl();
		long Bx = sc.nl();
		long By = sc.nl();
		long[] start = {sc.nl(),sc.nl()};
		long T = sc.nl();
		int N = 1;
		while (true) {
			long[] p = {Ax*points[N-1][0]+Bx,Ay*points[N-1][1]+By};
			if (dist(p,start) <= (long)(1e17)) {
				points[N] = p;
				N++;
			} else {
				break;
			}
		}
		//pw.println(Arrays.deepToString(points));
		int ans = 0;
		
		for (int i = 0; i < N; i++) {
			//go to i then go down to 0, then go up as high as possible.
			long dist = dist(start,points[i]);
			if (dist > T) continue;
			int pos = i;
			int val = 1;
			while (pos > 0) {
				dist += dist(points[pos],points[pos-1]);
				if (dist > T) break;
				pos--;
				val++;
			}
			if (dist > T || i == N-1) {
				ans = Math.max(ans,val);
				continue;
			}
			
			dist += dist(points[0],points[i+1]);
			if (dist > T) {
				ans = Math.max(ans,val);
				continue;
			}
			pos = i+1;
			val++;
			while (pos < N-1) {
				dist += dist(points[pos],points[pos+1]);
				if (dist > T) break;
				pos++;
				val++;
			}
			ans = Math.max(ans,val);
		}
		pw.println(ans);
		pw.close();
	}
	
	public static long dist(long[] p1, long[] p2) {
		return (Math.abs(p2[0]-p1[0])+Math.abs(p2[1]-p1[1]));
	}
	
	//Find the GCD of two numbers
	public static long gcd(long a, long b) {
		if (a < b) return gcd(b,a);
		if (b == 0)
			return a;
		else
			return gcd(b,a%b);
	}
	
	//Fast exponentiation (x^y mod m)
	public static long power(long x, long y, long m) { 
		long ans = 1;
		x %= m;
		while (y > 0) { 
			if(y % 2 == 1) 
				ans = (ans * x) % m; 
			y /= 2;  
			x = (x * x) % m;
		} 
		return ans; 
	}
	
	public static int[][] shuffle(int[][] array) {
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    int[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		return array;
	}
	
    public static int[][] sort(int[][] array) {
    	//Sort an array (immune to quicksort TLE)
 
		Arrays.sort(array, new Comparator<int[]>() {
			  @Override
        	  public int compare(int[] arr1, int[] arr2) {
				  return arr1[0]-arr2[0]; //ascending order
	          }
		});
		return array;
	}
    
    public static long[][] sort(long[][] array) {
    	//Sort an array (immune to quicksort TLE)
		Random rgen = new Random();
		for (int i = 0; i < array.length; i++) {
		    int randomPosition = rgen.nextInt(array.length);
		    long[] temp = array[i];
		    array[i] = array[randomPosition];
		    array[randomPosition] = temp;
		}
		Arrays.sort(array, new Comparator<long[]>() {
			  @Override
        	  public int compare(long[] arr1, long[] arr2) {
				  return 0;
	          }
		});
		return array;
	}
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int ni() { 
            return Integer.parseInt(next()); 
        } 
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str;
        }
    }
}