WOW Factor

Recall that string

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

Input

The input contains a single non-empty string

Output

Output a single integer, the wow factor of

Examples

input

Copy

vvvovvv

output

Copy

input

Copy

vvovooovovvovoovoovvvvovovvvov

output

Copy

Note

The first example is explained in the legend.

刚开始忘了两个int相加还是int，只用了一个longlong的sum，导致溢出。

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#include <xfunctional>
#define ll long long
#define mod 998244353
using namespace std;
int dir[4][2] = { {0,1},{0,-1},{-1,0},{1,0} };
const int maxn = 1e5 + 5;
const long long inf = 0x7f7f7f7f7f7f7f7f;

int main()
{
    string s;
    cin >> s;
    int v = 0, maxn = 0, initial = 0, firsto = 0;
    vector<ll> dp;
    for (int i = 0; s[i] == 'v'; i++)
        initial++;
    if (initial > 0)
        dp.push_back(initial - 1);
    else
        dp.push_back(0);
    for (int i = initial+1; i < s.size(); i++)
    {
        if (s[i] == 'v')
            maxn++;
        else
        {
            if (maxn > 1)
                dp.push_back(dp.back() + maxn - 1);
            else
                dp.push_back(dp.back());
            maxn = 0;
        }
        if (i == s.size() - 1 && maxn>1)
        {
            dp.push_back(dp.back() + maxn - 1);
        }
            
    }

    ll sum = 0;
    for (int i = 0; i < dp.size(); i++)
    {
        sum += (dp.back() - dp[i])*dp[i];
    }
    cout << sum;
    return 0;
}

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