cf 305A

E - Strange Addition

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 305A

Appoint description:  System Crawler  (2015-01-10)

Description

Unfortunately, Vasya can only sum pairs of integers (a, b), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.

Vasya has a set of k distinct non-negative integers d₁, d₂, ..., d_k.

Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?

Input

The first input line contains integer k(1 ≤ k ≤ 100) — the number of integers.

The second line contains k distinct space-separated integers d₁, d₂, ..., d_k(0 ≤ d_i ≤ 100).

Output

In the first line print a single integer n the maximum number of the chosen integers. In the second line print n distinct non-negative integers — the required integers.

If there are multiple solutions, print any of them. You can print the numbers in any order.

Sample Input

Input

4
100 10 1 0

Output

4
0 1 10 100 

Input

3
2 70 3

Output

2
2 70 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
using namespace std;
int k,a[110],b[110],t,j;
int main()
{
      scanf("%d",&k);
      for(int i=0;i<k;i++)
            scanf("%d",&a[i]);
      sort(a,a+k);
      int n=1;
      b[0]=a[0];
      for(int i=1;i<k;i++)
      {
            int x=a[i]%10;
            t=a[i]/10;
            int y=t%10;
            t=t/10;
            int z=t;
            for(j=0;j<n;j++)
            {
                  int x1=b[j]%10;
                  t=b[j]/10;
                  int y1=t%10;
                  t=t/10;
                  int z1=t;
                  if((x==0||x1==0)&&(y==0||y1==0)&&(z==0||z1==0))
                  {
                       continue;
                  }
                  else
                  {
                        break;
                  }
            }
            if(j==n)
            {
                  b[n++]=a[i];
            }
      }
      printf("%d\n",n);
      printf("%d",b[0]);
      for(int i=1;i<n;i++)
            printf(" %d",b[i]);
      printf("\n");
      return 0;
}