Problem:
Score : 300 points

Problem Statement

Given is a string S of length N-1.
Each character in S is < or >.

A sequence of N non-negative integers, a_1,a_2,\cdots,a_N, is said to be good when the following condition is satisfied for all i (1 \leq i \leq N-1):

If S_i= <: a_i<a_{i+1}

If S_i= >: a_i>a_{i+1}

Find the minimum possible sum of the elements of a good sequence of N non-negative integers.

Constraints

2 \leq N \leq 5 \times 10^5

S is a string of length N-1 consisting of < and >.

Input

Input is given from Standard Input in the following format:

S

Output

Find the minimum possible sum of the elements of a good sequence of N non-negative integers.

Sample Input 1

<>>

Sample Output 1

3

a=(0,2,1,0) is a good sequence whose sum is 3.
There is no good sequence whose sum is less than 3.

Sample Input 2

<>>><<><<<<<>>><

Sample Output 2

28

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdlib.h>
#include <string.h>
#include <stdio.h>

int main()
{   char s[500005];
    scanf("%s",s);
    int l=strlen(s);
    long long a[l+2];
    long long left=1,right=0,x=0;
    if(s[0]=='>'){
        int i=0;
        while(s[i]=='>'){
            i++;
        }x=i;
        for(i;i>=0;i--){
            a[x-i]=i;
        }
        x=i;
    }else{
        left=1;
    }
    for(int i=x;i<l;i++){
        if(s[i]==s[i+1]&&s[i]=='<'){
            left++;
        }
        if(s[i]==s[i+1]&&s[i]=='>'){
            right++;
        }
        if(s[i]=='<'&&s[i+1]=='>'){
            i++;
            while(s[i+right]=='>'){
                right++;
            }
                for(int j=0;j<=left;j++){
                    a[i-left+j]=j;
                }
                left=1;
                if(a[i]<right){
                    a[i]=right;
                    
                }right--;
                for(int j=0;j<=right;j++){
                    a[i+j+1]=right-j;
                }
                i+=right-1;
                right=0;
            }
            if(i==l-1&&s[i]=='<'){
                //left++;
                int k=0;
                i++;
                while(left>=0){
                    a[i-left]=k;
                    left--;
                    k++;
                }
            }
    }
    long long ans=0;
    for(int i=0;i<l+1;i++){
        ans+=a[i];
        //printf("%d ",a[i]);
    }
    printf("%lld",ans);
    return 0;
}