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 <image> 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.
/**
 * Created by ankurverma1994
 */

import java.io.*;
import java.util.*;
import java.math.*;

public class Prob5 {
    final int mod = (int) 1e9 + 7;
    final double eps = 1e-6;
    final double pi = Math.PI;

    //------------> Solution starts here!!
    @SuppressWarnings("Main Logic")
    void solve() {
        n = ii();
        m = ii();
        long ans = 0;
        for (int i = 1; i <= n; i++) {
            int a = i % 5;
            if (a == 0) a = 5;
            ans += getNum(a);
//            out.println(getNum(a));
        }
        out.println(ans);
    }

    int n, m;

    int getNum(int a) {
        if (a == 1) {
            if (m >= 4) return (m - 4) / 5 + 1;
            return 0;
        }
        if (a == 2) {
            if (m >= 3) return (m - 3) / 5 + 1;
            return 0;
        }
        if (a == 3) {
            if (m >= 2) return (m - 2) / 5 + 1;
            return 0;
        }
        if (a == 4) {
            if (m >= 1) return (m - 1) / 5 + 1;
            return 0;
        }
        return m / 5 ;
    }

    //------------> Solution ends here!!
    InputStream obj;
    PrintWriter out;
    String check = "";

    public static void main(String[] args) throws IOException {
        new Thread(null, new Runnable() {
            public void run() {
                try {
                    new Prob5().main1();
                } catch (IOException e) {
                    e.printStackTrace();
                } catch (StackOverflowError e) {
                    System.out.println("RTE");
                }
            }
        }, "1", 1 << 26).start();
    }

    void main1() throws IOException {
        out = new PrintWriter(System.out);
        obj = check.isEmpty() ? System.in : new ByteArrayInputStream(check.getBytes());
        // obj=check.isEmpty() ? new FileInputStream("/home/ankurverma1994/input.txt") : new ByteArrayInputStream(check.getBytes());
        solve();
        out.flush();
        out.close();
    }

    byte inbuffer[] = new byte[1024];
    int lenbuffer = 0, ptrbuffer = 0;

    int readByte() {
        if (lenbuffer == -1) throw new InputMismatchException();
        if (ptrbuffer >= lenbuffer) {
            ptrbuffer = 0;
            try {
                lenbuffer = obj.read(inbuffer);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }
        if (lenbuffer <= 0) return -1;
        return inbuffer[ptrbuffer++];
    }

    String is() {
        int b = skip();
        StringBuilder sb = new StringBuilder();
        while (!(isSpaceChar(b))) // when nextLine, (isSpaceChar(b) && b!=' ')
        {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    int ii() {
        int num = 0, b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    long il() {
        long num = 0;
        int b;
        boolean minus = false;
        while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        while (true) {
            if (b >= '0' && b <= '9') {
                num = num * 10 + (b - '0');
            } else {
                return minus ? -num : num;
            }
            b = readByte();
        }
    }

    boolean isSpaceChar(int c) {
        return (!(c >= 33 && c <= 126));
    }

    int skip() {
        int b;
        while ((b = readByte()) != -1 && isSpaceChar(b)) ;
        return b;
    }

    float nf() {
        return Float.parseFloat(is());
    }

    double id() {
        return Double.parseDouble(is());
    }

    char ic() {
        return (char) skip();
    }

    int[] iia(int n) {
        int a[] = new int[n];
        for (int i = 0; i < n; i++) a[i] = ii();
        return a;
    }

    long[] ila(int n) {
        long a[] = new long[n];
        for (int i = 0; i < n; i++) a[i] = il();
        return a;
    }

    String[] isa(int n) {
        String a[] = new String[n];
        for (int i = 0; i < n; i++) a[i] = is();
        return a;
    }

    double[][] idm(int n, int m) {
        double a[][] = new double[n][m];
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) a[i][j] = id();
        return a;
    }

    int[][] iim(int n, int m) {
        int a[][] = new int[n][m];
        for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) a[i][j] = ii();
        return a;
    }
}
