CodeForces - 1117C

You a captain of a ship. Initially you are standing in a point (x1,y1)(x1,y1) (obviously, all positions in the sea can be described by cartesian plane) and you want to travel to a point (x2,y2)(x2,y2).

You know the weather forecast — the string ss of length nn, 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 s1s1, the second day — s2s2, the nn-th day — snsn and (n+1)(n+1)-th day — s1s1 again and so on.

Ship coordinates change the following way:

• if wind blows the direction U, then the ship moves from (x,y)(x,y)to (x,y+1)(x,y+1);
• if wind blows the direction D, then the ship moves from (x,y)(x,y)to (x,y−1)(x,y−1);
• if wind blows the direction L, then the ship moves from (x,y)(x,y)to (x−1,y)(x−1,y);
• if wind blows the direction R, then the ship moves from (x,y)(x,y)to (x+1,y)(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)(x,y) it will move to the point (x−1,y+1)(x−1,y+1), and if it goes the direction U, then it will move to the point (x,y+2)(x,y+2).

You task is to determine the minimal number of days required for the ship to reach the point (x2,y2)(x2,y2).

Input

The first line contains two integers x1,y1x1,y1 (0≤x1,y1≤1090≤x1,y1≤109) — the initial coordinates of the ship.

The second line contains two integers x2,y2x2,y2 (0≤x2,y2≤1090≤x2,y2≤109) — 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 nn (1≤n≤1051≤n≤105) — the length of the string ss.

The fourth line contains the string ss 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 (x2,y2)(x2,y2).

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

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

In the third example the ship can never reach the point (0,1)(0,1).

思路很简单：二分天数，然后check下是否可以。可以向下二分，不可以向上二分。

天数上界调大点就行了。

 

不过有一点求大佬告知！！就是我注释的代码和 下面的for循环操作  性质上是一样的，但结果WA在了8.。

就下面这组数据，不知道为啥，是编译原理的问题吗？？？？求解啊啊啊！！！

 

更新：  发现问题了，注释下的代码没更新sx[i],sy[i]数组，（只更新了一个方向上的，这竟然神奇的过了7个数据）；

	#include<cstdio>
	#include<iostream>
	#include<algorithm>
	#define maxn 100000+100
	typedef long long ll;
	using namespace std;
	char s[maxn];
	ll sumx[maxn],sumy[maxn];
	ll las(ll a)
	{
		if(a>0)
		return a;
		return -a;
	}
	int main()
	{
		ll x1,x2,y1,y2,len;
		cin>>x1>>y1>>x2>>y2>>len;
		cin>>s;
		for(int i=1;i<=len;++i)
		{
			sumx[i]=sumx[i-1];
			sumy[i]=sumy[i-1];
			if(s[i-1]=='U')
				++sumy[i];
			else
				if(s[i-1]=='D')
					--sumy[i];
				else
					if(s[i-1]=='L')
						--sumx[i];
					else
						++sumx[i];
		}
	/*	sumx[0]=0;
		sumy[0]=0;
		ll sx=0,sy=0 ;
		for(int i=1;i<=len;++i)
		{
			
			if(s[i-1]=='U')
				sumy[i]=++sy;
			else
				if(s[i-1]=='D')
					sumy[i]=--sy;
				else
					if(s[i-1]=='L')
						sumx[i]=--sx;
					else
						sumx[i]=++sx;
		}*/ 
	//	printf("%lld--%lld\n",sx[len],sy[len]);
		ll l=1,r=1e18,nn,ans=-1,nx,ny;
		while(l<=r)
		{
			ll mid=(l+r)/2;//mid为二分出的天数 
			nx=x1+mid/len*sumx[len]+sumx[mid%len];
			ny=y1+mid/len*sumy[len]+sumy[mid%len];
			nn=las(x2-nx)+las(y2-ny);
			if(mid>=nn)
			{
				r=mid-1;
				ans=mid;
			}
			else
			{
				l=mid+1;
			}
		}
		cout<<ans<<endl;
		return 0;
	}