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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int count_right(int N, vector<char> S){
    int count = 0;
    int i = N;
    while(true){
        if(S.at(i) == '>')
            count++;
        else
            break;
        i++;        
    }

    return count;
}

int count_left(int N, vector<char> S){
    int count = 0;
    int i = N;
    while(true){
        if(S.at(i) == '<')
            count++;
        else
            break;
        i--;
    }

    return count;
}

void good_numbers(vector<int> &a, vector<char> S, int N){
    for(int i = 0; i < N-1; i++){
        if(S.at(0) == '>'){
            int right_number = count_right(0, S);
            a.at(0) = right_number;
            for(int j = 1; j <= right_number; j++){
                a.at(j) = right_number - j;
            }
        }

        if(S.at(N-2) == '<'){
            int left_number = count_left(N-2, S);
            a.at(N-1) = left_number;
            for(int j = 1; j <= left_number; j++){
                a.at(N-1-j) = left_number - j;
            }
        }

        if(S.at(i) == '<' && S.at(i+1) == '>'){
            int left_number = count_left(i, S);
            int right_number = count_right(i+1, S);

            int max_num = max(left_number, right_number);

            a.at(i+1) = max_num;

            if(max_num == left_number){
                for(int j = 1; j <= left_number; j++){
                    a.at(i+1-j) = left_number - j;
                }

                for(int j = 1; j <= right_number; j++){
                    if(j == 1){
                        a.at(i+1+j) = max_num - right_number;
                    }
                    else{
                        a.at(i+1+j) = right_number - j;
                    }
                }
            }

            else{
                for(int j = 1; j <= left_number; j++){
                    if(j == 1){
                        a.at(i+1-j) = max_num - left_number;
                    }
                    else{
                        a.at(i+1-j) = left_number - j;
                    }
                }
                for(int j = 1; j <= right_number; j++){
                    a.at(i+1+j) = right_number - j;
                }
            }
        }
    }
}

 
int main() {
    string str;
    cin >> str;
    int N = str.size() + 1;
    vector<char> S(N-1);
    for(int i = 0; i < N-1; i++){
        S.at(i) = str.at(i);
    }

    vector<int> a(N);

    good_numbers(a, S, N);

    for(int i = 0; i < N; i++){
        cout << a.at(i) << endl;
    }

    int sum = 0;
    for(int i = 0; i < N; i++){
        sum += a.at(i);
    }

    cout << sum << endl;
    return 0;
}