F. The Neutral Zone
time limit per test
5 seconds
memory limit per test
16 megabytes
input
standard input
output
standard output

Notice: unusual memory limit!

After the war, destroyed cities in the neutral zone were restored. And children went back to school.

The war changed the world, as well as education. In those hard days, a new math concept was created.

As we all know, logarithm function can be described as: $$\log(p_1^{a_1}p_2^{a_2}...p_k^{a_2}) = a_1 \log p_1 + a_2 \log p_2 + ... + a_k \log p_k$$ Where $p_1^{a_1}p_2^{a_2}...p_k^{a_2}$ is the prime factorization of a integer. A problem is that the function uses itself in the definition. That is why it is hard to calculate.

So, the mathematicians from the neutral zone invented this: $$\text{exlog}_f(p_1^{a_1}p_2^{a_2}...p_k^{a_2}) = a_1 f(p_1) + a_2 f(p_2) + ... + a_k f(p_k)$$

Notice that $\text{exlog}_f(1)$ is always equal to $0$.

This concept for any function $f$ was too hard for children. So teachers told them that $f$ can only be a polynomial of degree no more than $3$ in daily uses (i.e., $f(x) = Ax^3+Bx^2+Cx+D$).

"Class is over! Don't forget to do your homework!" Here it is: $$\sum_{i=1}^n \text{exlog}_f(i)$$

Help children to do their homework. Since the value can be very big, you need to find the answer modulo $2^{32}$.

Input

The only line contains five integers $n$, $A$, $B$, $C$, and $D$ ($1 \le n \le 3 \cdot 10^8$, $0 \le A,B,C,D \le 10^6$).

Output

Print the answer modulo $2^{32}$.

Examples
Input
12 0 0 1 0
Output
63
Input
4 1 2 3 4
Output
136
Note

In the first sample:

$\text{exlog}_f(1) = 0$

$\text{exlog}_f(2) = 2$

$\text{exlog}_f(3) = 3$

$\text{exlog}_f(4) = 2 + 2 = 4$

$\text{exlog}_f(5) = 5$

$\text{exlog}_f(6) = 2 + 3 = 5$

$\text{exlog}_f(7) = 7$

$\text{exlog}_f(8) = 2 + 2 + 2 = 6$

$\text{exlog}_f(9) = 3 + 3 = 6$

$\text{exlog}_f(10) = 2 + 5 = 7$

$\text{exlog}_f(11) = 11$

$\text{exlog}_f(12) = 2 + 2 + 3 = 7$

$\sum_{i=1}^{12} \text{exlog}_f(i)=63$

In the second sample:

$\text{exlog}_f(1) = 0$

$\text{exlog}_f(2) = (1 \times 2^3 + 2 \times 2^2 + 3 \times 2 + 4) = 26$

$\text{exlog}_f(3) = (1 \times 3^3 + 2 \times 3^2 + 3 \times 3 + 4) = 58$

$\text{exlog}_f(4) = 2 \times \text{exlog}_f(2) = 52$

$\sum_{i=1}^4 \text{exlog}_f(i)=0+26+58+52=136$