C. Vasya and Robot

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has got a robot which is situated on an infinite Cartesian plane, initially in the cell (0,0)(0,0). Robot can perform the following four kinds of operations:

• U — move from (x,y)(x,y) to (x,y+1)(x,y+1);
• D — move from (x,y)(x,y) to (x,y−1)(x,y−1);
• L — move from (x,y)(x,y) to (x−1,y)(x−1,y);
• R — move from (x,y)(x,y) to (x+1,y)(x+1,y).

Vasya also has got a sequence of nn operations. Vasya wants to modify this sequence so after performing it the robot will end up in (x,y)(x,y).

Vasya wants to change the sequence so the length of changed subsegment is minimum possible. This length can be calculated as follows: maxID−minID+1maxID−minID+1, where maxIDmaxID is the maximum index of a changed operation, and minIDminID is the minimum index of a changed operation. For example, if Vasya changes RRRRRRR to RLRRLRL, then the operations with indices 22, 55 and 77 are changed, so the length of changed subsegment is 7−2+1=67−2+1=6. Another example: if Vasya changes DDDD to DDRD, then the length of changed subsegment is 11.

If there are no changes, then the length of changed subsegment is 00. Changing an operation means replacing it with some operation (possibly the same); Vasya can't insert new operations into the sequence or remove them.

Help Vasya! Tell him the minimum length of subsegment that he needs to change so that the robot will go from (0,0)(0,0) to (x,y)(x,y), or tell him that it's impossible.

Input

The first line contains one integer number n (1≤n≤2⋅105)n (1≤n≤2⋅105) — the number of operations.

The second line contains the sequence of operations — a string of nn characters. Each character is either U, D, L or R.

The third line contains two integers x,y (−109≤x,y≤109)x,y (−109≤x,y≤109) — the coordinates of the cell where the robot should end its path.

Output

Print one integer — the minimum possible length of subsegment that can be changed so the resulting sequence of operations moves the robot from (0,0)(0,0) to (x,y)(x,y). If this change is impossible, print −1−1.

Examples

input

Copy

5
RURUU
-2 3

output

Copy

3

input

Copy

4
RULR
1 1

output

Copy

0

input

Copy

3
UUU
100 100

output

Copy

-1

Note

In the first example the sequence can be changed to LULUU. So the length of the changed subsegment is 3−1+1=33−1+1=3.

In the second example the given sequence already leads the robot to (x,y)(x,y), so the length of the changed subsegment is 00.

In the third example the robot can't end his path in the cell (x,y)(x,y).

這道題今日最佳，出的好，自己做的也不錯，雖然最後也沒A，但就差一個二分了。

題意：給你一個操作字串，包含四個操作，修改字串最短長度的區間能到從（0，0）到（x，y）。

分析：

這題最大的簡便：x和y的累加量很好計算

1、明確一點，操作的先後順序是不影響結果。

2、選定一個修改區間長度，區間裡操作可以任意修改，因為題意要求最短區間長度，不是修改運算元量

3、我們一開始可以想到區間長度從len從0到n開始暴力列舉，我們可以先把長度為len的區間扣下來，計算剩下的累計值，這裡需要一個字首和的思想，假設區間長度為len，起點為i，終點為i+len-1，需要減去的就是t=a[i+len-1]-a[i-1],剩下的為t2=a[n]-t即可，然後t2與剩餘的步數比較，如果滿足t2==step，正好，但是，如果step<=t2,需要滿足(len-step)%2==0，因為存在LR，轉移的來回

4.然後二分是我沒想到的，我以為這是一個遞增的肯定從中間遍歷，但是正因為它是遞增的，所以才正好用二分。就差這一步啊。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
const int M=200005;
int n;
int x[M],y[M];
int ex,ey;
int cal(int len)
{
    for(int i=1;i+len<=n;i++)
    {
        int j=i+len;
        int t1=x[j]-x[i-1];
        int t2=y[j]-y[i-1];
        int tx=x[n]-t1;
        int ty=y[n]-t2;
        int step;
        //cout<<len<<" "<<i<<" "<<step<<endl;
        step=abs(tx-ex)+abs(ty-ey);
        if(len+1>=step&&(len+1-step)%2==0)
        {
            return 1;
        }
    }
    return 0;
}
int main()
{
    scanf("%d",&n);
    string s;
    cin>>s;
    for(int i=0;i<n;i++)
    {
      if(s[i]=='U')
        y[i+1]=1;
      else if(s[i]=='D')
        y[i+1]=-1;
      else if(s[i]=='R')
        x[i+1]=1;
      else if(s[i]=='L')
        x[i+1]=-1;
    }


    scanf("%d%d",&ex,&ey);
    int m=abs(ex)+abs(ey);
    if(n<m)
    {
        printf("-1");
        return 0;
    }


    for(int i=2;i<=n;i++)
    {
      y[i]=y[i]+y[i-1];
      x[i]=x[i]+x[i-1];
    }
    if(y[n]==ey&&x[n]==ex)
    {
        printf("0");
        return 0;
    }
    int ans=-1;
    int l=0,r=n-1;
    while(l<=r)
    {
       int mid=(l+r)/2;
       if(cal(mid))
       {
           ans=mid+1;
           r=mid-1;
       }
       else
       {
           l=mid+1;
       }
    }
    printf("%d",ans);
    return 0;
}