Problem: 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: $$$ exlog_f(p_1^{a_1}p_2^{a_2}...p_k^{a_2}) = a_1 f(p_1) + a_2 f(p_2) + ... + a_k f(p_k) $$$

Notice that 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: $$$ ∑_{i=1}^n exlog_f(i) $$$

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 ≤ n ≤ 3 ⋅ 10^8, 0 ≤ A,B,C,D ≤ 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:

exlog_f(1) = 0

exlog_f(2) = 2

exlog_f(3) = 3

exlog_f(4) = 2 + 2 = 4

exlog_f(5) = 5

exlog_f(6) = 2 + 3 = 5

exlog_f(7) = 7

exlog_f(8) = 2 + 2 + 2 = 6

exlog_f(9) = 3 + 3 = 6

exlog_f(10) = 2 + 5 = 7

exlog_f(11) = 11

exlog_f(12) = 2 + 2 + 3 = 7

 ∑_{i=1}^{12} exlog_f(i)=63 

In the second sample:

exlog_f(1) = 0

exlog_f(2) = (1 × 2^3 + 2 × 2^2 + 3 × 2 + 4) = 26

exlog_f(3) = (1 × 3^3 + 2 × 3^2 + 3 × 3 + 4) = 58

exlog_f(4) = 2 × exlog_f(2) = 52

 ∑_{i=1}^4 exlog_f(i)=0+26+58+52=136 

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Vadim Semenov
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskF solver = new TaskF();
        solver.solve(1, in, out);
        out.close();
    }

    static final class TaskF {
        private int n;
        private int a;
        private int b;
        private int c;
        private int d;

        public void solve(int __, InputReader in, PrintWriter out) {
            n = in.nextInt();
            a = in.nextInt();
            b = in.nextInt();
            c = in.nextInt();
            d = in.nextInt();
            int sqrt = 3;
            while (sqrt * sqrt < n) {
                sqrt++;
            }
            boolean[] isPrime = new boolean[sqrt + 1];
            Arrays.fill(isPrime, 2, isPrime.length, true);
            int primesQty = 0;
            for (int p = 2; p < isPrime.length; ++p) {
                if (isPrime[p]) {
                    primesQty++;
                    for (int np = p * p; np < isPrime.length; np += p) {
                        isPrime[np] = false;
                    }
                }
            }
            int[] primes = new int[primesQty];
            primesQty = 0;
            int answer = 0;
            for (int prime = 2; prime < isPrime.length; ++prime) {
                if (isPrime[prime]) {
                    primes[primesQty++] = prime;
                    answer += g(prime);
                }
            }
            for (int block = 1; block <= n / sqrt; ++block) {
                Arrays.fill(isPrime, true);
                int from = block * isPrime.length;
                int to = Math.min(n + 1, from + isPrime.length);
                for (int prime : primes) {
                    for (int np = (from + prime - 1) / prime * prime; np < to; np += prime) {
                        isPrime[np - from] = false;
                    }
                }
                for (int prime = from; prime < to; ++prime) {
                    if (isPrime[prime - from]) {
                        answer += g(prime);
                    }
                }
            }
            out.println(Integer.toUnsignedString(answer));
        }

        private int g(int x) {
            int result = 0;
            int fx = f(x);
            for (int pow = x; ; pow *= x) {
                result += n / pow * fx;
                if (n / x < pow) break;
            }
            return result;
        }

        private int f(int x) {
            return d + x * (c + x * (b + x * a));
        }

    }

    static class InputReader {
        private final BufferedReader reader;
        private StringTokenizer tokenizer;

        public InputReader(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(readLine());
            }
            return tokenizer.nextToken();
        }

        public String readLine() {
            String line;
            try {
                line = reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }

    }
}