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"

using namespace std;

using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using vvvll = vector<vector<vector<ll>>>;
using vvvvll = vector<vector<vector<vector<ll>>>>;
using pl4 = pair<ll,ll>;
using str = string;
using vpl4 = vector<pair<ll,ll>>;

#define sz size()
#define be begin()
#define en end()
#define fi first
#define se second
#define pb push_back
#define mp make_pair

#define llin(x) ll (x);cin >>(x);
#define stin(x) str (x);cin >>(x);
#define vllin(x,n) vll (x)(n);FOR(i,0,n-1){cin >>(x)[i];}
#define vllin2(a,b,n) vll (a)(n);vll (b)(n);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i];}
#define vllin3(a,b,c,n) vll (a)(n);vll (b)(n);vll (c)(n);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i]>>(c)[i];}
#define vlling(x,n) (x).assign(n,0);FOR(i,0,n-1){cin >>(x)[i];}
#define vlling2(a,b,n) (a).assign(n,0);(b).assign(n,0);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i];}
#define vlling3(a,b,c,n) (a).assign(n,0);(b).assign(n,0);(c).assign(n,0);FOR(i,0,n-1){cin >>(a)[i]>>(b)[i]>>(c)[i];}
#define vpl4in(x,n) vpl4 (x)((n),mp(0,0));FOR(i,0,n-1){cin >>x[i].fi>>x[i].se;}
#define FOR(i,a,b) for(ll i = a ; i <= b ; i++)
#define rFOR(i,b,a) for(ll i = a; i >= b ; i--)
#define SORT(x) sort(x.be, x.en)
#define rSORT(x) sort(x.rbegin(), x.rend())
#define say(x) cout<<(x);
#define sal(x) cout<<(x)<<endl;
#define says(x) cout<<(x)<<(' ');
#define sas cout<<(' ');
#define sayR(x) cout<<fixed<<setprecision(10)<<(x);
#define salR(x) cout<<fixed<<setprecision(10)<<(x)<<endl;
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define Imp(a) cout <<((a)?"Possible":"Impossible")<<endl;
#define IMP(a) cout <<((a)?"POSSIBLE":"IMPOSSIBLE")<<endl;
#define pow(a,b) ll(pow(a,b))

ll MOD=1000000007;

signed main(){
    stin(s);
    ll t=0;
    ll k=0;
    ll c=0;
    FOR(i,0,s.sz-1){
        if(s[i]=='>'){
            c++;
            k+=c;
            if(c>=t) k++;
        }
        if(s[i]=='<'){
            if(c>=0) t=0;
            t++;
            c=-1;
            k+=t;
        }
    }
    sal(k);
}