Codeforces Round #358 (Div. 2) A Alyona and Numbers（水题）

思路：(x+y)%5==0，余数的组合可以有(0,0),(1,4),(2,3),(3,2),(4,1)五种，那么分别计算出来就可以了、

#include<bits\stdc++.h>
using namespace std;
#define LL long long
LL num1[6];
LL num2[6];
int main()
{
LL n,m;
cin >> n >> m;
for(int i = 0;i<5;i++)
{
num1[i]=n/5;
if(n%5>=i)
num1[i]++;
num2[i]=m/5;
if(m%5>=i)
num2[i]++;
}
num1[0]--;
num2[0]--;
LL ans = 0;
ans+=num1[0]*num2[0]+num1[1]*num2[4]+num1[2]*num2[3]+num1[3]*num2[2]+num1[4]*num2[1];
cout << ans << endl;
}


A. Alyona and Numbers

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

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.