Given N integers,  output the number of different pairs that two numbers have common digits. For example, given 3 integers: 1 11 21, there are 3 common digit pairs: <1, 11>, <1, 21>, <11, 21>.

The first line has a positive integer N (1 ≤ N ≤ 1,000,000), then follow N different positive integers in the next N lines, each integer is no more than 10¹⁸.

Output the numbers of different common digit pairs.

#include <stdio.h>

char s[][];

int main()

{int n;

scanf("%d",&n);

getchar();

for(int i=;i<n;i++)

scanf("%s",s[i]);

int e=,f;

for(int i=;i<n;i++)

for(int j=i+;j<n;j++){

    f=;

    for(int k=;s[i][k];k++){

    for(int l=;s[j][l];l++)

    if(s[i][k]==s[j][l]){

        e++;f=;

        break;

    }

    if(f)break;}

}

    printf("%d",e);

    return ;

}

#include <stdio.h>

__int64 s[];

int a[];

int main()

{

    int n;

    a[]=;

    for(int i=; i<; i++)

        a[i]=a[i-]*;

    scanf("%d",&n);

    getchar();

    while(n--)

    {

        char c;

        bool b[]= {};

        while(c=getchar(),c!='\n')

        {

            b[c-]=;

        }

        int e=;

        for(int k=; k<; k++)

            if(b[k])

                e+=a[k];

        s[e]++;

    }

    __int64 f=;

    for(int i=; i<; i++)

    {

        f+=s[i]*(s[i]-)/;

        for(int j=; j<i; j++)

            if(i&j)

                f+=s[i]*s[j];

    }

    printf("%I64d",f);

    return ;

}