After finishing eating her bun, Alyona came up with two integers n and m. She decided to write down two columns of integers —

the first column containing integers from 1 to n and the second containing integers from 1 to m. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.

Formally, Alyona wants to count the number of pairs of integers (x, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and  equals 0.

As usual, Alyona has some troubles and asks you to help.

Input

The only line of the input contains two integers n and m (1 ≤ n, m ≤ 1 000 000).

Output

Print the only integer — the number of pairs of integers (x

Examples

Input

6 12

Output

14

Input

11 14

Output

31

Input

1 5

Output

1

Input

3 8

Output

5

Input

5 7

Output

7

Input

21 21

Output

88

Note

Following pairs are suitable in the first sample case:

• for x
   = 1 fits y equal to 4 or 9;
• for x = 2 fits y equal to 3 or 8;
• for x = 3 fits y equal to 2, 7 or 12;
• for x = 4 fits y equal to 1, 6 or 11;
• for x = 5 fits y equal to 5 or 10;
• for x = 6 fits y equal to 4 or 9.

Only the pair (1, 4) is suitable in the third sample case.

題意：給兩個數，n,m,這連個數分別從1-n,1-m，從兩個數中取倆個數相加，和是5的倍數；

思路：既然是5的倍數，那，我們每次從n這個數中取數只要只要他%5的餘數就行了，再加上m，（表示最多可以從m中取這麼大的數），再/5就數要求的，從m中取出可以和%5組成5的倍數的個數了，

具體看程式碼就理解，數比較大，要用long long 才能AC，我一開始就卡在這，頭疼。。。。

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
int main()
{
    long long int n,m;
    scanf("%lld%lld",&n,&m);
   long long  int ans=0;
    int t;
    for(int i=1; i<=n; i++)
    {t=i%5;
 ans+=(t+m)/5;
    }
    printf("%lld\n",ans);
    return 0;
}