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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let s: Vec<char> = iterator.next().unwrap().chars().collect();

    let mut sum: Vec<usize> = vec![0; s.len() + 1];
    for i in 1..(s.len() + 1) {
        sum[i] = sum[i - 1] + i;
    }

    let mut parts: Vec<String> = Vec::new();
    let mut part = String::new();
    let mut prev: char = ' ';
    for &c in &s {
        if c != prev {
            if prev != ' ' {
                parts.push(part.clone());
            }
            part = String::new();
        }
        part.push(c);
        prev = c;
    }
    parts.push(part);

    let mut ans = 0;
    for i in 0..parts.len() {
        let part = &parts[i];
        if part.starts_with('<') {
            ans += sum[part.len()];
        } else {
            if i == 0 {
                ans += sum[part.len()];
            } else if part.len() <= parts[i - 1].len() {
                ans += sum[part.len() - 1];
            } else {
                ans -= parts[i - 1].len();
                ans += sum[part.len()];
            }
        }
    }
    return ans.to_string();
}