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: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
int main(void){
    // Your code here!
    char str[500000];
    int seq[500001];
    scanf("%s", str);
    int count = 0;
    int upcount = 0;
    int downcount = 0;
    int i = 0;
    int sum = 0;
    
    for(i=0; i<=strlen(str); i++){
        seq[i]=0;
    }
    
    for(i=strlen(str)-1; i>=0; i--){
        if(str[i] == '>') {
            if(downcount == 0){
                seq[i+1] = 0;
                seq[i] = 1;
                downcount++;
                if(upcount > 0){
                    upcount = 0;
                }
            } else {
                seq[i] = seq[i+1] + 1;
                downcount++;
            }
        }
    }
    for(i=0; i<strlen(str); i++){
        if(str[i] == '<'){
            if(upcount == 0){
                seq[i] = 0;
                if(seq[i+1]==0)
                    seq[i+1] = 1;
                upcount++;
                if(downcount > 0){
                    downcount = 0;
                }
            } else {
                if(seq[i+1] < seq[i] + 1) 
                    seq[i+1] = seq[i] + 1;
                upcount++;
            }
        }
        sum += seq[i];
    }
    sum += seq[strlen(str)];
    printf("%d", sum);
    return 0;
}