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)

#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = s; i < (int)(n); i++)
#define Clear(a) a = decltype(a)();
typedef long long ll;
typedef vector<vector<ll>> vvl;
typedef vector<ll> vl;
typedef vector<vector<char>> vvc;
typedef vector<char> vc;
typedef pair<ll,ll> pll;
typedef priority_queue<string> priquestring;
typedef priority_queue<int> priqueint;
typedef priority_queue<ll> priquell;
typedef priority_queue<char> priquechar;
typedef priority_queue<string,vector<string>,greater<string>> rpriquestring;
typedef priority_queue<int,vector<int>,greater<int>> rpriqueint;
typedef priority_queue<ll,vector<ll>,greater<ll>> rpriquell;
typedef priority_queue<char,vector<char>,greater<char>> rpriquechar;
ll max(ll x,ll y){
if(x>y)return x;
else return y;
}
ll min(ll x,ll y){
  return x+y-max(x,y);
}

const ll big=1000000007;
int exp(int x,int y){
    if(y==0)return 1;//0^0=1
    if(x==1||x==0)return x;
    if(y%2==1)return (exp(x,y-1)*x)%big;
    int t=exp(x,y/2);
    return (t*t)%big;
}
void chmin(ll a,ll b){
    if(a>b)a=b;
    return;
}

const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};


int main(){
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cout << fixed << setprecision(10);
    /*--------------------------------*/
    
    //>>>>> <<<< >>>>>>>>> <<
    //     0    (?)       0
    //<<<<がk個,そのあと>>>>がm個→k(k-1)/2+max(k,m)+m(m-1)/2
    //<<<<>>>>>が途切れた段階でcntしてやれば良い
    ll ans=0;
    ll left=0;ll right=0;
    string s;cin >> s;
    int n=(int)s.size()+1;
    rep(i,n-1){
        if(i==n-2){
            if(s[i]=='>'){
                right++;
                ans+=left*(left-1)/2+max(right,left)+right*(right-1)/2;
            }else{
                left++;
                ans+=left*(left-1)/2;
            }
            continue;
        }
        if(s[i]=='>'&&s[i+1]=='<'){
            right++;
            ans+=left*(left-1)/2+max(left,right)+right*(right-1)/2;
            left=0;right=0;
        }else if(s[i]=='>'&&s[i+1]=='>')right++;
        else if(s[i]=='<')left++;
    }
    cout << ans << '\n';
}