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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    var s = readLine()!!

    var prev = '-'
    var greater = 0
    var smaller = 0

    val ss = mutableListOf<Int>()

    s.forEach {
        if (it == '<') {
            if (prev == '-' || prev== '<') {
                greater++
            } else {
                ss.add(smaller)
                greater++
                smaller = 0
            }
            prev = '<'
        } else {
            if (prev == '-' || prev== '>') {
                smaller--
            } else {
                ss.add(greater)
                smaller--
                greater = 0
            }
            prev = '>'
        }
    }

    if (greater > 0) {
        ss.add(greater)
    }
    if (smaller < 0) {
        ss.add(smaller)
    }

    var ans = 0
    var last = 0
    for (i in ss.indices) {
        if (ss[i] > 0) {   // '<'
            repeat(ss[i]-1) {
                ans += (it+1)
            }
            last = ss[i]-1

            if (i == ss.size-1) {
                repeat(ss[i]) {
                    ans += it+1
                }
            }

        } else {    // '>'
            if (last < -ss[i]) {
                repeat(-ss[i]) {
                    ans += (it+1)
                }
            } else {
                ans += last+1
                repeat(-ss[i]-1) {
                    ans += (it+1)
                }
            }
        }
    }

    println(ans)
}