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)

#pragma GCC optimize("O3")
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

using ll = long long;
using P = pair<int, int>;
using T = tuple<int, int, int>;

template <class T> inline T chmax(T &a, const T b) {return a = (a < b) ? b : a;}
template <class T> inline T chmin(T &a, const T b) {return a = (a > b) ? b : a;}

constexpr int MOD = 1e9 + 7;
constexpr int inf = 1e9;
constexpr long long INF = 1e18;
constexpr double pi = acos(-1);
constexpr double EPS = 1e-10;

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    string s; cin>>s;
    int n = s.size();

    // from left
    vector<int> l(n+1, 0);
    for(int i=0; i<n; i++){
        int cnt = 0;
        while(i < n && s[i] == '<'){
            l[i] = cnt, l[i+1] = cnt + 1;
            i++, cnt++;
        }
    }

    // from right
    vector<int> r(n+1, 0);
    for(int i=n-1; i>=0; i--){
        int cnt = 0;
        while(i >= 0 && s[i] == '>'){
            r[i] = cnt + 1, r[i+1] = cnt;
            i--, cnt++;
        }
    }

    vector<ll> ans(n+1);
    for(int i=0; i<=n; i++){
        ans[i] = max(l[i], r[i]);
    }

    cout << accumulate(ans.begin(), ans.end(), 0LL) << endl;

    return 0;
}