You a captain of a ship. Initially you are standing in a point (x_1, y_1) (obviously, all positions in the sea can be described by cartesian plane) and you want to travel to a point (x_2, y_2). 

You know the weather forecast — the string s of length n, consisting only of letters U, D, L and R. The letter corresponds to a direction of wind. Moreover, the forecast is periodic, e.g. the first day wind blows to the side s_1, the second day — s_2, the n-th day — s_n and (n+1)-th day — s_1 again and so on. 

Ship coordinates change the following way:

  * if wind blows the direction U, then the ship moves from (x, y) to (x, y + 1); 
  * if wind blows the direction D, then the ship moves from (x, y) to (x, y - 1); 
  * if wind blows the direction L, then the ship moves from (x, y) to (x - 1, y); 
  * if wind blows the direction R, then the ship moves from (x, y) to (x + 1, y). 



The ship can also either go one of the four directions or stay in place each day. If it goes then it's exactly 1 unit of distance. Transpositions of the ship and the wind add up. If the ship stays in place, then only the direction of wind counts. For example, if wind blows the direction U and the ship moves the direction L, then from point (x, y) it will move to the point (x - 1, y + 1), and if it goes the direction U, then it will move to the point (x, y + 2).

You task is to determine the minimal number of days required for the ship to reach the point (x_2, y_2).

Input

The first line contains two integers x_1, y_1 (0 ≤ x_1, y_1 ≤ 10^9) — the initial coordinates of the ship.

The second line contains two integers x_2, y_2 (0 ≤ x_2, y_2 ≤ 10^9) — the coordinates of the destination point.

It is guaranteed that the initial coordinates and destination point coordinates are different.

The third line contains a single integer n (1 ≤ n ≤ 10^5) — the length of the string s.

The fourth line contains the string s itself, consisting only of letters U, D, L and R.

Output

The only line should contain the minimal number of days required for the ship to reach the point (x_2, y_2).

If it's impossible then print "-1".

Examples

Input


0 0
4 6
3
UUU


Output


5


Input


0 3
0 0
3
UDD


Output


3


Input


0 0
0 1
1
L


Output


-1

Note

In the first example the ship should perform the following sequence of moves: "RRRRU". Then its coordinates will change accordingly: (0, 0) → (1, 1) → (2, 2) → (3, 3) → (4, 4) → (4, 6).

In the second example the ship should perform the following sequence of moves: "DD" (the third day it should stay in place). Then its coordinates will change accordingly: (0, 3) → (0, 3) → (0, 1) → (0, 0).

In the third example the ship can never reach the point (0, 1).
import java.util.*;
import java.math.*;

// **** C. Magic Ship ****

public class C {
	static char [] in = new char [100000];
	public static void main (String [] arg) throws Throwable {
		int x1 = nextInt();
		int y1 = nextInt();
		int x2 = nextInt();
		int y2 = nextInt();
		long dX = x2 - x1;
		long dY = y2 - y1;	
		
		int n = nextInt();
		char [] s = next().toCharArray();
		if (s.length != n) System.err.println("???");
		if (dX == 0 && dY == 0) {System.out.println(0); return;}
		long wX = 0;
		long wY = 0;
		long [] wXa = new long [n];
		long [] wYa = new long [n];
		for (int i =0; i<s.length; ++i) {
			if (s[i] == 'U') wY++;
			if (s[i] == 'D') wY--;
			if (s[i] == 'R') wX++;
			if (s[i] == 'L') wX--;
			wYa[i] = wY;
			wXa[i] = wX;
		}
		// Binary search
		long MAX = 2 * n * ((long)Integer.MAX_VALUE);
		long OLD_MAX = MAX;
		long MIN = 0;
		BigInteger OVERFLOW = BigInteger.valueOf(Long.MAX_VALUE);
		while (MIN < MAX) {
			long PIV = (MIN + MAX) / 2;
			long DIV = PIV / n;
			int REM = (int)(PIV % n);
			
			long Xpos = wX * DIV;
			long Ypos = wY * DIV;
			BigInteger XposB = BigInteger.valueOf(wX).multiply(BigInteger.valueOf(DIV));
			BigInteger YposB = BigInteger.valueOf(wY).multiply(BigInteger.valueOf(DIV));
			if (XposB.compareTo(OVERFLOW) > 0 || YposB.compareTo(OVERFLOW) > 0) {
				MAX = PIV-1;
			}
			
			if (REM > 0) {
				Xpos += wXa[REM-1];
				Ypos += wYa[REM-1];
			}
			long Xgoal = dX - Xpos;
			long Ygoal = dY - Ypos;
			if (Math.abs(Xgoal) + Math.abs(Ygoal) <= PIV) {
				//System.err.println("Can Do " + PIV + " , " + Xgoal + " , " + Ygoal);
				// Can do it
				MAX = PIV;
			} else {
				//System.err.println("Cant do " + PIV + " , " + Xgoal + " , " + Ygoal);
				MIN = PIV+1;
			}
		}
		System.out.println((OLD_MAX == MAX) ? -1 : MAX);
	}

	
	
	
	
	
	
	
	
	
	
	
	
	
	
	/************** FAST IO CODE FOLLOWS *****************/
	public static long nextLong() throws Throwable {
		long i = System.in.read();boolean neg = false;while (i < 45) i = System.in.read();if (i == 45) {neg=true;i=48;}i = i - 48;
		int j = System.in.read();while (j > 32) {i*=10;i+=j-48;j = System.in.read();}return (neg) ? -i : i;
	}
	public static int nextInt() throws Throwable {return (int)nextLong();}
	public static String next() throws Throwable {
		int i = 0; while (i < 42 && i != -1) i = System.in.read(); int cptr = 0; while (i >= 42) { in[cptr++] = (char)i; i = System.in.read();}
		return new String(in, 0,cptr);
	}
	/**** LIBRARIES ****/
	public static long gcdL(long a, long b) {while (b != 0) {long tmp = b;b = (a % b);a = tmp;}return a;}
	public static int gcd(int a, int b) {while (b != 0) {int tmp = b;b = (a % b);a = tmp;}return a;}
}

/* Full Problem Text:
  
You a captain of a ship.
Initially you are standing in a point (x_1, y_1) (obviously, all positions in the sea can be described by cartesian plane) and you want to travel to a point (x_2, y_2).
You know the weather forecast  the string s of length n, consisting only of letters U, D, L and R.
The letter corresponds to a direction of wind.
Moreover, the forecast is periodic, e.g. the first day wind blows to the side s_1, the second day  s_2, the n-th day  s_n and (n+1)-th day  s_1 again and so on.
Ship coordinates change the following way:
 
 if wind blows the direction U, then the ship moves from (x, y) to (x, y + 1); 
 if wind blows the direction D, then the ship moves from (x, y) to (x, y - 1); 
 if wind blows the direction L, then the ship moves from (x, y) to (x - 1, y); 
 if wind blows the direction R, then the ship moves from (x, y) to (x + 1, y).
The ship can also either go one of the four directions or stay in place each day.
If it goes then it's exactly 1 unit of distance.
Transpositions of the ship and the wind add up.
If the ship stays in place, then only the direction of wind counts.
For example, if wind blows the direction U and the ship moves the direction L, then from point (x, y) it will move to the point (x - 1, y + 1),
and if it goes the direction U, then it will move to the point (x, y + 2).
You task is to determine the minimal number of days required for the ship to reach the point (x_2, y_2).

 */