CF | Alyona and Numbers

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, y) such that 1 ≤ x ≤ n, 1 ≤ y ≤ m and (x + y) is divisible by 5.

Example

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.

 #include<bits/stdc++.h>
 using namespace std;

 int main()
 {
     int n,m,t,count1,count2,temp;
     while(cin>>n>>m) {
         long long int ans = ;
         if(n > m) {
             t = n;
             n = m;
             m = t;
         }
         count1 = m / ; // 求出有多少组 （1 2 3 4 5）
         count2 = m % ; // 求出剩余多少个数 

         for(int i = ; i <= n; i++)
         {
             ans += count1; //每一个数在一个循环组有能够找到一个相加模5等于0的数
             temp = i % ;  //temp i 在 循环组的位置
             if((-temp )<= count2 && count2 !=  )  //根据循环组的内规律进行匹配  1--4 2--3
             {
                 ans++;
             }
         }

         cout<<ans<<endl;
     }

     return ;
 }