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<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INF 10000000000
#define mod 1000000007
typedef long long int ll;
ll u(const void *a, const void *b){return *(ll*)a-*(ll*)b;}
ll d(const void *a, const void *b){return *(ll*)b-*(ll*)a;}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll gcd(ll a,ll b){
  if(!b)return a;
  return gcd(b,a%b);
}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
ll f(ll a){
  return a*(a+1)/2;
}
int main(void){
  char s[500001];
  scanf("%s",s);
  ll a[500000],flg=(s[0]=='>'),cnt1=1,cnt2=0,ans=0;
  FOR(strlen(s)-1){
    if(s[i]-s[i+1]){
      a[cnt2]=cnt1;
      cnt2++;
      cnt1=1;
    }else{
      cnt1++;
    }
  }
  a[cnt2]=cnt1;
  cnt2++;
  a[cnt2]=1;
  if(flg){
    ans+=f(a[0])+f(a[cnt2-1]);
    cnt2-=2;
  }
  for(ll i=0;i<cnt2;i+=2){
    if(flg){
      i++;
      flg--;
    }
    ans+=f(max(a[i],a[i+1]))+f(min(a[i],a[i+1])-1);
  }
  PRL(ans);
  return 0;
}