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;
#define int long long
#define endl '\n'
#define vi vector<int>
#define vb vector<bool>
#define pii pair<int,int>
#define mod 1000000007
int add(int n){
    return (n*(n+1))/2;
}
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    string s;
    cin>>s;
    int n = s.size(),i,sum=0;
    vi v;
    v.push_back(0);
    for(i=0;i<n-1;i++){
        if((s[i]=='>' && s[i+1]=='<') || (s[i]=='<' && s[i+1]=='>'))v.push_back(i+1);
    }
    v.push_back(n);
    if(s[0]=='<' && s[n-1]=='>'){
        for(i=0;i<v.size()-2;i+=2){
            sum+=add(max((v[i+1]-v[i]),(v[i+2]-v[i+1])))+add(min((v[i+1]-v[i]),(v[i+2]-v[i+1]))-1);
        }
    }
    if(s[0]=='<' && s[n-1]=='<'){
        for(i=0;i<v.size()-3;i+=2){
            sum+=add(max((v[i+1]-v[i]),(v[i+2]-v[i+1])))+add(min((v[i+1]-v[i]),(v[i+2]-v[i+1]))-1);
        }
        sum+=add(v[v.size()-1]-v[v.size()-2]);
    }
    if(s[0]=='>' && s[n-1]=='>'){
        for(i=1;i<v.size()-2;i+=2){
            sum+=add(max((v[i+1]-v[i]),(v[i+2]-v[i+1])))+add(min((v[i+1]-v[i]),(v[i+2]-v[i+1]))-1);
        }
        sum+=add(v[1]-v[0]);
    }
    if(s[0]=='>' && s[n-1]=='<'){
        for(i=1;i<v.size()-3;i+=2){
            sum+=add(max((v[i+1]-v[i]),(v[i+2]-v[i+1])))+add(min((v[i+1]-v[i]),(v[i+2]-v[i+1]))-1);
        }
        sum+=add(v[v.size()-1]-v[v.size()-2]);
        sum+=add(v[1]-v[0]);   
    }
    cout<<sum;
    return 0;
}