687B: Remainders Game

Codeforces Round #360 Editorial [+ Challenges!]

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, ..., c_n 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, ..., c_n (1 ≤ c_i ≤ 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.

Hint

Assume the answer of a test is No. There must exist a pair of integers x₁ and x₂ such that both of them have the same remainders after dividing by any c_i, but they differ in remainders after dividing by k. Find more facts about x₁ and x₂!

Solution

Consider the x₁ and x₂ from the hint part. We have x₁ - x₂ ≡ 0 () for each 1 ≤ i ≤ n.

So:

We also have (). As a result:

We've found a necessary condition. And I have to tell you it's also sufficient!

Assume , we are going to prove there exists x₁, x₂ such that x₁ - x₂ ≡ 0 () (for each 1 ≤ i ≤ n), and ().

A possible solution is x₁ = lcm(c₁, c₂, ..., c_n) and x₂ = 2 × lcm(c₁, c₂, ..., c_n), so the sufficiency is also proved.

So you have to check if lcm(c₁, c₂, ..., c_n) is divisible by k, which could be done using prime factorization of k and c_i values.

For each integer x smaller than MAXC, find it's greatest prime divisor gpd_x using sieve of Eratosthenes in .

Then using gpd array, you can write the value of each coin as p₁^q₁p₂^q₂...p_m^q_m where p_i is a prime integer and 1 ≤ q_i holds. This could be done in by moving from c_i to and adding gpd_ci to the answer. And you can factorize k by the same way. Now for every prime p that , see if there exists any coin i that the power of p in the factorization of c_i is not smaller than the power of p in the factorization of k.

Complexity is .

C++ code

//     . .. ... .... ..... be name khoda ..... .... ... .. .     \

#include <bits/stdc++.h>
using namespace std;

inline int in() { int x; scanf("%d", &x); return x; }
const long long N = 1200021;

int cntP[N], isP[N];

int main()
{
	for(int i = 2; i < N; i++)
		if(!isP[i])
			for(int j = i; j < N; j += i)
				isP[j] = i;
	int n = in(), k = in();
	for(int i = 0; i < n; i++)
	{
		int x = in();
		while(x > 1)
		{
			int p = isP[x];
			int cnt = 0;
			while(x % p == 0)
			{
				cnt++;
				x /= p;
			}
			cntP[p] = max(cntP[p], cnt);
		}
	}
	bool ok = 1;
	while(k > 1)
	{
		ok &= (cntP[isP[k]] > 0);
		cntP[isP[k]]--;
		k /= isP[k];
	}
	cout << (ok ? "Yes
" : "No
");
}

原文链接：Codeforces Round #360 Editorial [+ Challenges!] - Codeforces