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: Wrong Answer)

/*Author: ADITYA SHETH*/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const double PI=acos(-1.0);
#define t1(x)             cerr<<#x<<"="<<x<<endl
#define t2(x, y)          cerr<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define t3(x, y, z)       cerr<<#x<<"=" <<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define t4(a,b,c,d)       cerr<<#a<<"="<<a<<" "<<#b<<"="<<b<<" "<<#c<<"="<<c<<" "<<#d<<"="<<d<<endl
#define t5(a,b,c,d,e)     cerr<<#a<<"="<<a<<" "<<#b<<"="<<b<<" "<<#c<<"="<<c<<" "<<#d<<"="<<d<<" "<<#e<<"="<<e<<endl
#define t6(a,b,c,d,e,f)   cerr<<#a<<"="<<a<<" "<<#b<<"="<<b<<" "<<#c<<"="<<c<<" "<<#d<<"="<<d<<" "<<#e<<"="<<e<<" "<<#f<<"="<<f<<endl
#define GET_MACRO(_1,_2,_3,_4,_5,_6,NAME,...) NAME
#define t(...) GET_MACRO(__VA_ARGS__,t6,t5, t4, t3, t2, t1)(__VA_ARGS__)
//freopen("output.txt","w",stdout);
//freopen("input.txt","r",stdin);
/*-------------------------------------------------------------------------------------------------------------------------------------*/
#define MOD 1000000007
#define endl "\n" 
#define int long long // remove when constraints are tight.
#define inf 1e18
/*-------------------------------------------------------------------------------------------------------------------------------------*/

int32_t main()
{
	ios::sync_with_stdio(0);cin.tie(0);
	string s;
	cin>>s;
	int n=s.length();
	int ans=0;
	int ind=0;
	while(ind<n && s[ind]!='>')ind++;
	ans+=((ind)*(ind-1))/2;
	int cur=0;
	int cnt =1;
	char opt[]={'>','<'};
	for(int i=ind;i<n;)
	{
		while(i<n && s[i]==opt[cur])
		{
			cnt++;
			i++;
		}
		if(cur==1)
		{
			ans+=(cnt*(cnt+1))/2;
		}
		else
		{
			ans+=(cnt*(cnt-1))/2;
		}
		cnt=0;
		cur=!cur;
	}

	cout<<ans<<endl;
}