CodeForces 682A Alyona and Numbers

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.

_(:з」∠)_我真应该多做点思维题了。我一开始居然试图傻不拉唧的暴力……然后果然TLE了，TLE在第十组数据上。啊对了……后来数据类型搞错了，还WA了一次。

这个题方法很巧，把1~n,1~m范围内的数字的余数的大小求出来，然后用余数相加等于5或者0的数字进行排列……

话说我还看到有人能达到O(5)……哎……怎么能想得出来呢……

#include<stdio.h>
#include<string.h>
__int64 a[10],b[10];

int main()
{
int i,n,m;
while(~scanf("%d%d",&n,&m))
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(i=1;i<=n;i++)
a[i%5]++;
for(i=1;i<=m;i++)
b[i%5]++;
__int64 ans=a[0]*b[0]+a[1]*b[4]+a[2]*b[3]+a[3]*b[2]+a[4]*b[1];
printf("%I64d\n",ans);
}
return 0;
}