Codeforces 626A Robot Sequence(模拟)

A. Robot Sequence

time limit per test：2 seconds

memory limit per test：256 megabytes

input：standard input

output：standard output

Calvin the robot lies in an infinite rectangular grid. Calvin's source code contains a list of n commands, each either 'U', 'R', 'D', or 'L' — instructions to move a single square up, right, down, or left, respectively. How many ways can Calvin execute a non-empty contiguous substrings of commands and return to the same square he starts in? Two substrings are considered different if they have different starting or ending indices.

Input

The first line of the input contains a single positive integer, n (1 ≤ n ≤ 200) — the number of commands.

The next line contains n characters, each either 'U', 'R', 'D', or 'L' — Calvin's source code.

Output

Print a single integer — the number of contiguous substrings that Calvin can execute and return to his starting square.

Examples

Input

6
URLLDR

Output

2

Input

4
DLUU

Output

0

Input

7
RLRLRLR

Output

12

Note

In the first case, the entire source code works, as well as the "RL" substring in the second and third characters.

Note that, in the third case, the substring "LR" appears three times, and is therefore counted three times to the total result.

题目链接：http://codeforces.com/contest/626/problem/A

题意：在一块方格里，给出一串指令包含U，D，L，R这些字母，分别表示上下左右移动一格。问有多少个子串（包含自身）能使得物体回到出发位置。

分析：模拟一下找子串的个数即可！

解释下样例3：子串为：RL，RLRL，RLRLRL，LR，LRLR，LRLRLR，RL，RLRL，LR，LRLR，RL，LR，共12条，所以输出为12

下面给出AC代码：

 #include <bits/stdc++.h>
 using namespace std;
 inline int read()
 {
     int x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch='-')
             f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 int n;
 char s[];
 int main()
 {
     n=read();
     cin>>s;
     int e=;
     for(int i=;i<n;i++)
     {
         int a=,b=,c=,d=;
         for(int j=i;j<n;j++)
         {
             if(s[j]=='U')
                 a++;
             if(s[j]=='D')
                 b++;
             if(s[j]=='L')
                 c++;
             if(s[j]=='R')
                 d++;
             if(a==b&&c==d)
                 e++;
         }
     }
     cout<<e<<endl;
     return ;
 }