CF-499div2-E-裴蜀定理

E. Border

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Astronaut Natasha arrived on Mars. She knows that the Martians are very poor aliens. To ensure a better life for the Mars citizens, their emperor decided to take tax from every tourist who visited the planet. Natasha is the inhabitant of Earth, therefore she had to pay the tax to enter the territory of Mars.

There are $\mathrm{nn banknote\; denominations\; on\; Mars:\; the\; value\; of ii-th\; banknote\; is aiai.\; Natasha\; has\; an\; infinite\; number\; of\; banknotes\; of\; each\; denomination.}$

Martians have $\mathrm{kk fingers\; on\; their\; hands,\; so\; they\; use\; a\; number\; system\; with\; base kk.\; In\; addition,\; the\; Martians\; consider\; the\; digit dd (in\; the\; number\; system\; with\; base kk)\; divine.\; Thus,\; if\; the\; last\; digit\; in\; Natasha\text{'}s\; tax\; amount\; written\; in\; the\; number\; system\; with\; the\; base kk is dd,\; the\; Martians\; will\; be\; happy.\; Unfortunately,\; Natasha\; does\; not\; know\; the\; Martians\text{'}\; divine\; digit\; yet.}$

Determine for which values $\mathrm{dd Natasha\; can\; make\; the\; Martians\; happy.}$

Natasha can use only her banknotes. Martians don't give her change.

Input

The first line contains two integers $\mathrm{nn and kk (1\le n\le 1000001\le n\le 100000, 2\le k\le 1000002\le k\le 100000) —\; the\; number\; of\; denominations\; of\; banknotes\; and\; the\; base\; of\; the\; number\; system\; on\; Mars.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109) —\; denominations\; of\; banknotes\; on\; Mars.}$

All numbers are given in decimal notation.

Output

On the first line output the number of values $\mathrm{dd for\; which\; Natasha\; can\; make\; the\; Martians\; happy.}$

In the second line, output all these values in increasing order.

Print all numbers in decimal notation.

Examples

input

Copy

2 8
12 20

output

Copy

2
0 4 

input

Copy

3 10
10 20 30

output

Copy

1
0 

Note

Consider the first test case. It uses the octal number system.

If you take one banknote with the value of $\mathrm{1212,\; you\; will\; get 148148 in\; octal\; system.\; The\; last\; digit\; is 4848.}$

If you take one banknote with the value of $\mathrm{1212 and\; one\; banknote\; with\; the\; value\; of 2020,\; the\; total\; value\; will\; be 3232.\; In\; the\; octal\; system,\; it\; is 408408.\; The\; last\; digit\; is 0808.}$

If you take two banknotes with the value of $\mathrm{2020,\; the\; total\; value\; will\; be 4040,\; this\; is 508508 in\; the\; octal\; system.\; The\; last\; digit\; is 0808.}$

No other digits other than ${\mathrm{0808 and 4848 can\; be\; obtained.\; Digits 0808 and 4848 could\; also\; be\; obtained\; in\; other\; ways.}}^{}$

The second test case uses the decimal number system. The nominals of all banknotes end with zero, so Natasha can give the Martians only the amount whose decimal notation also ends with zero.

　　给出n种钞票的面值(十进制下),每种钞票数量inf，问在k进制下能组合成的所有面额的最低位的数有哪些。

这就等价于是  a1*x1+a2*x2+......+an*xn=g  ,g就是组合成的钞票面值，由裴蜀定理可知d=y*gcd(a1,a2,,,an),

我们只要求出gcd(a1,a2,,,,,,an)，然后把所有情况(y<k)扫一下就好了。

#include<bits/stdc++.h>
using namespace std;
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
set<long long >S;
int main(){
    long long  a,n,k,i,j,g=0;
    cin>>n>>k;
    for(i=1;i<=n;++i){
        cin>>a;
        g=gcd(g,a);
    }
    for(i=0,j=0;j<=k;j++,i+=g) S.insert(i%k);
    cout<<S.size()<<endl;
    for(set<long long >::iterator it=S.begin();it!=S.end();++it){
        printf("%lld%c",*it,it==S.end()?'
':' ');
    }
    return 0;
}