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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define sim template <class c
#define ris return *this
#define dor > debug &operator<<
#define eni(x)                                                                        \
	sim > typename enable_if<sizeof dud<c>(0) x 1, debug &>::type operator<<(c i) \
	{
sim > struct rge
{
	c b, e;
};
sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c *x) -> decltype(cerr << *x, 0);
sim > char dud(...);
struct debug
{
#ifdef LOCAL
	~debug()
	{
		cerr << endl;
	}
	eni(!=) cerr << boolalpha << i;
	ris;
} eni(==) ris << range(begin(i), end(i));
}
sim, class b dor(pair<b, c> d)
{
	ris << "(" << d.first << ", " << d.second << ")";
}
sim dor(rge<c> d)
{
	*this << "[";
	for (auto it = d.b; it != d.e; ++it)
		*this << ", " + 2 * (it == d.b) << *it;
	ris << "]";
}
#else
	sim dor(const c &)
	{
		ris;
	}
#endif
}
;
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

#define forn(i, n) for (ll i = 0; i < int(n); i++)
#define rfor(n, i) for (ll i = (n - 1); i >= 0; --i)
#define all(V) V.begin(), V.end()
#define rall(V) V.rbegin(), V.rend()
#define len(V) (int)V.size()
#define ll long long
#define ld long double
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define endl '\n'
#define M_PI 3.14159265358979323846
#define EPS 1e-7

void solve()
{	
	string s; cin >> s;
	int n = s.length();
	ll res = 0, cnt = 0;
	for (int i = 0; i < n; ++i)
	{
		if (s[i] == '>') 
		{
			cnt++;
			res += cnt;
		}
		else 
		{
			cnt = 0;
		}
	}
	cnt = 0;
	for (int i = n - 1; i >= 0; --i)
	{
		if (s[i] == '<')
		{
			cnt++;
			res += cnt;
		}
		else
		{
			cnt = 0;
		}
	}
	cnt = 0;
	//cout << res << endl;
	for (int i = 1; i < n; ++i)
	{
		if (s[i] != s[i - 1])
		{
			res--;
			i++;
		}
	}
	cout << res;
}

int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	solve();
}