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)

#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
#define pa pair<int, int>
#define pal pair<long long, long long>
#define pali pair<long long, int>
#define pad pair<double, double>
#define pb push_back
#define mp make_pair
#define COUT(v)                              \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cout << v.at(i) << endl;             \
    }
#define REP(i, n) for (int64_t i = 0; i < n; ++i)
#define FOR(i, r, n) for (int64_t i = (r); i < n; ++i)
#define VIN(v)                               \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cin >> (v).at(i);                    \
    }
 
typedef vector<bool> bvec;
typedef vector<int> ivec;
typedef vector<long long> lvec;
typedef vector<double> dvec;
typedef vector<pa> pavec;
typedef vector<pali> palivec;
typedef vector<pal> palvec;
 
typedef vector<vector<bool> > bmat;
typedef vector<vector<int> > imat;
typedef vector<vector<long long> > lmat;
 
typedef vector<string> svec;
typedef vector<vector<string> > smat;
const ll infll = (1LL << 60) - 1;
const int inf = (1 << 30) - 1;
const ll MOD = 1000000007;
 
ll gcd(ll x, ll y)
{
    ll r = x % y;
    if (r == 0)
        return y;
    else
        return gcd(y, r);
}
 
ll lcm(ll x, ll y)
{
    return x * y / gcd(x, y);
}
 
lvec mfactor(ll n)
{
    bvec ip(n, true);
    lvec mf(n, -1);
    ip[0] = false;
    ip[1] = false;
    mf[0] = 0;
    mf[1] = 1;
    REP(i, n)
    {
        if (ip[i])
        {
            mf[i] = i;
            for (ll j = i * i; j < n; j += i)
            {
                ip[j] = false;
                if (mf[j] == -1)
                    mf[j] = i;
            }
        }
    }
    return mf;
}
 
palivec get_prime(ll n, const lvec &mf)
{
    palivec plist;
    while (n != 1)
    {
        int cnt = 0;
        ll m = mf[n];
        while (mf[n] == m)
        {
            cnt++;
            n /= m;
        }
        plist.pb(pali(m, cnt));
    }
    return plist;
}
 
void COMinit(int m, lvec &fac, lvec &finv)
{
    lvec inv(m);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < m; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
 
    }
}
 
ll pow_mod(ll a, ll n)
{
    ll x = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            x = x * a % MOD;
        }
        a = a * a % MOD;
        n >>= 1;
    }
    return x;
}
 
 
ll COM(int n, int k, const lvec &fac, const lvec &finv)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return (fac[n] * (finv[k] * finv[n - k] % MOD)) % MOD;
}
 
ll modinv(ll a, ll m)
{
    ll b = m, u = 1, v = 0;
    while (b)
    {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}


int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s;
    cin >> s;
    int n = s.size();
    ivec a(n+1, -1);
    int cntl=0, cntr=0;
    ivec al(n+1), ar(n+1);
    al[0] = 0;
    ar[n] = 0;
    for (int i=1; i<n+1; i++){
        if(s[i-1] == '<') cntl++;
        else cntl=0;
        al[i] = cntl;
    }
    
    for (int i=n-1; i>=0; i--){
        if(s[i] == '>') cntr++;
        else cntr=0;
        ar[i] = cntr;
    }
    ll ans=0;
    for (int i=0; i<n+1; i++){
        //cout << al[i] << ar[i] << endl;
        ans += max(al[i], ar[i]);
    }
    cout << ans << endl;
}