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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cstdio>
#include <complex>
#include <numeric>
#include <string.h>
#include <random>
#include <cmath>
#define rep(i,n) for (int i = 0; i < (int)n; i++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<pi, pi> pp;
typedef pair<ll, ll> pl;
double PI = 3.141592653589793238462643383279;
const double EPS = 1e-9;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const int inf = 1 << 30;
const ll linf = 1LL << 60;

string str;
int cl[500010], cr[500001];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> str;
    int si = str.size();
    rep(i,str.size()) {
        if (i) cl[i] = cl[i-1];
        else cl[i] = 0;
        if (str[i] == '<') {
            cl[i]++;
        } else {
            cl[i] = 0;
        }
    }
    for (int i = si-1; i >= 0; i--) {
        cr[i] = cr[i+1];
        if (str[i] == '>') cr[i]++;
        else cr[i] = 0;
    }
    ll ans = 0;
    rep(i,si+1) {
        int now = 0;
        now = cr[i];
        if (i != 0) {
            now = max(now, cl[i-1]);
        }
        ans += now;
    }
    cout << ans << endl;
}