The number 151 is a prime palindrome because it is both a prime number and a palindrome (it is the same number when read forward as backward). Write a program that finds all prime palindromes in the range of two supplied numbers a and b (5 <= a < b <= 1000,000,000); both a and b are considered to be within the range .

Input

Line 1: Two integers, a and b

Output

The list of palindromic primes in numerical order, one per line.

Sample Input

5 500

Sample Output

5

7

11

101

131

151

181

191

313

353

373

383

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int i;
int Is_Prime(int a)
{
    double t = a;
    for(i=2;i<=sqrt(t);i++)
    {
        if(a % i == 0)
            return 0;
    }
        return 1;
}
int diglen(int c)
{
    if(c/10 == 0)
        return 1;
    int count=0;
    while(c)
    {
        c=c/10;
        count++;
    }
    return count;
}

void IntoStr(int a,char* str) //將數字轉換為字符串
{
    int h= diglen(a)-1,i=0;
    while(h+1)
    {
    int k,c,d=1;
    for(k=0;k<h;k++)
    {
        d=d*10;
    }

    c=a/d;//取本次數字的最高位
    str[i++] = c+48;
    a = a%d; //取余數
    h--;
    }
    str[i]='\0'; //最后末尾加字符串結束符
}

int Is_Pal(int b)
{
    int len = diglen(b);
    int ok = 1,j;
    char* str = (char*)malloc(sizeof(char)*(len+1));
    IntoStr(b,str); //將數字轉換為字符數組，也可以用itoa函數或 sprintf函數
    for(i=0,j=len-1;i < len/2;i++,j--)
    {
        if(str[i] != str[j])
            ok = 0;
        
    }
    free(str);
    if(ok == 1)
        return 1;
    else
        return 0;
}

int main()
{
    int a,b,j;
    scanf("%d%d",&a,&b);
    for(j=a;j<=b;j++)
    {
        if(Is_Prime(j) && Is_Pal(j))
            printf("%d\n",j);
    }
    return 0;
}