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: Accepted)

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <algorithm>
#include <functional>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <tuple>
#include <numeric>
#define ll long long
using namespace std;

ll B(ll x) {
    return (x*(x+1))/2;
}

ll Solve(const vector<pair<int,int>>& A) {
    int minval = 0;
    ll ans = 0;
    int N=A.size();
    for(int i=0;i<N;i++) {
        if (A[i].first==0) { // <
            int ct = A[i].second;
            //printf("<%d\n",ct);

            ans += B(ct-1);
            minval += ct;
        } else {
            int ct = A[i].second;
            //printf(">%d\n",ct);

            minval = max(minval, A[i].second);
            //printf("minval=%d %lld\n",minval,minval + B(ct-1));
            ans += minval + B(ct-1);
            minval = 0;
        }
    }
    ans += minval;
    return ans;
}


int main() {
    string S;
    cin>>S;
    
    vector<pair<int,int>> cts;
    for(int i=0;i<S.size();) {
        int j=i;
        while(j+1<S.size() && S[j+1]==S[i]) {
            j++;
        }
        cts.push_back({S[i]=='>',j-i+1});
        i=j+1;
    }
    
    

    cout << Solve(cts) << endl;


}