Codeforces 687B - Remainders Game (剩余定理)

B. Remainders Game

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Today Pari and Arya are playing a game called Remainders.

Pari chooses two positive integer x and k,
and tells Arya k but not x.
Arya have to find the value 

.
There are n ancient numbers c1, c2, ..., cn and
Pari has to tell Arya 

 if
Arya wants. Given k and the ancient values, tell us if Arya has a winning strategy independent of value of x or
not. Formally, is it true that Arya can understand the value 

 for
any positive integer x?

Note, that 

 means
the remainder of x after dividing it by y.

Input

The first line of the input contains two integers n and k (1 ≤ n,  k ≤ 1 000 000) —
the number of ancient integers and value k that is chosen by Pari.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 1 000 000).

Output

Print "Yes" (without quotes) if Arya has a winning strategy independent of value of x,
or "No" (without quotes) otherwise.

Examples

input

4 5
2 3 5 12

output

Yes

input

2 7
2 3

output

No

Note

In the first sample, Arya can understand 

 because 5 is
one of the ancient numbers.

In the second sample, Arya can't be sure what 

 is.
For example 1 and 7 have the
same remainders after dividing by 2 and 3,
but they differ in remainders after dividing by 7.

 通过中国剩余定理得知x%m1=a1,x%m2=a2...x%mn=an 若a1~an已知,且gcd(mi,mj)=1,令M=πmi,则x%k1*M是可以确定的(因为x的全部解为(k*M+sum(ai*Mi*Mi^(-1))))，当k1=1时即为题目所问。所以只需要判断k的所有因数是否包含在πci中,即求lcm(ci,ci+1)(i<n)可不可以被k整除。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
#define F(x,a,b) for (int x=a;x<=b;i++)
#define me(x) memset(x,0,sizeof(a))
#define MAXN 10100000
#define ll long long
int gcd(int a,int b) {return b?gcd(b,a%b):a;}
int a[MAXN];
int main()
{
int n,m;ll k;
scanf("%d%d",&n,&m);
F(i,1,n)scanf("%d",&a[i]);
k=a[1];F(i,2,n) k%a[i]?k=k/gcd(k,a[i])*a[i]%m:k=k*(a[i]/gcd(k,a[i]))%m;
k%m?cout<<"No":cout<<"Yes";
}