B. WOW Factor

Recall that string aa is a subsequence of a string bb if aa can be obtained from bb by deletion of several (possibly zero or all) characters. For example, for the string aa="wowwo", the following strings are subsequences: "wowwo", "wowo", "oo", "wow", "", and others, but the following are not subsequences: "owoo", "owwwo", "ooo".

The wow factor of a string is the number of its subsequences equal to the word "wow". Bob wants to write a string that has a large wow factor. However, the "w" key on his keyboard is broken, so he types two "v"s instead.

Little did he realise that he may have introduced more "w"s than he thought. Consider for instance the string "ww". Bob would type it as "vvvv", but this string actually contains three occurrences of "w":

  • "vvvv"
  • "vvvv"
  • "vvvv"

For example, the wow factor of the word "vvvovvv" equals to four because there are four wows:

  • "vvvovvv"
  • "vvvovvv"
  • "vvvovvv"
  • "vvvovvv"

Note that the subsequence "vvvovvv" does not count towards the wow factor, as the "v"s have to be consecutive.

For a given string ss, compute and output its wow factor. Note that it is not guaranteed that it is possible to get ss from another string replacing "w" with "vv". For example, ss can be equal to "vov".

Input

The input contains a single non-empty string ss, consisting only of characters "v" and "o". The length of ss is at most 106106.

Output

Output a single integer, the wow factor of ss.

Examples

input

vvvovvv

output

4

input

vvovooovovvovoovoovvvvovovvvov

output

100

题意：两个连续的v可以组合成w，问字符串a的有多少个可以组成 wow 的子串

计算o之前w的个数 l，和o的个数 t，那么之后的一个“w”可以与之前的wo组成 l*t 个wow，也就是累加 l 次的 t

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))

int main()
{
	string a;
	int i;
	ll s=0,t=0,l=0;
	cin>>a;
	for(i=1;i<a.length();i++)
	{
		if(a[i]=='v'&&a[i-1]=='v')
		{
			s+=t;
			l++; 
		}
		if(a[i]=='o')
			t+=l;
	}
	cout<<s<<endl;
	return 0;
}