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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b){return *(int*)a - *(int*)b;} /* 1,2,3,4.. */
int des(const void *a, const void *b){return *(int*)b - *(int*)a;} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b){return *(char*)a - *(char*)b;} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b){return strcmp(*(const char **)a, *(const char **)b);} /* aaa,aab.. */
#define min(a,b) (a < b ? a: b)
#define max(a,b) (a > b ? a: b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 500001
#define MOD 1000000007

typedef long long int lli;

long long int Sum(int x) {
  return (1 + x) * x / 2;;
}

int main(void) {
  char s[MAX];
  scanf("%s", s);

  int n = strlen(s);

  long long int vsize = 0, val[MAX]; // <,>,...,<,>となる配列
  char now = '<';
  int cnt = 0;
  rep(i, 0, n) {
    if (s[i] == now) cnt++;
    else {
      val[vsize++] = cnt;
      now = now == '<' ? '>' : '<';
      cnt = 1;
    }
  }
  val[vsize++] = cnt;

  if (vsize % 2 == 0) val[vsize++] = 0;

  long long int ans = 0;
  for (int i = 0; i < vsize; i += 2) {
    if (val[i] > val[i + 1]) 
      ans += Sum(val[i]) + Sum(val[i + 1] - 1);
    else
      ans += Sum(val[i] - 1) + Sum(val[i + 1]);
  }
  printf("%lld\n", ans);
  return 0;
}