Bear and Poker CodeForces - 574C

B - Bear and Poker

CodeForces - 574C

     

Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are n players (including Limak himself) and right now all of them have bids on the table. i-th of them has bid with size a_i dollars.

Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?

Input

First line of input contains an integer n (2 ≤ n ≤ 10⁵), the number of players.

The second line contains n integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the bids of players.

Output

Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.

Example

Input

4
75 150 75 50

Output

Yes

Input

3
100 150 250

Output

No

Note

In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.

It can be shown that in the second sample test there is no way to make all bids equal.

题意，给出一列数，能否使某些数只乘2 和3，使的全相等。

只要，每个数除去它们的gcd后，全是2^x* 3 ^y，那一定，可以* 2 * 3变为，2 ^(max(x)) * 3 ^(max(y)).一开始想复杂了导致超时，没想到只要简单这么一改基本差不多久可以code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[100100];
int n;
int gcd(int a,int b){
    if(b==0)return a;
    else return gcd(b,a%b);
}
int main(){
    int i,j;
    scanf("%d",&n);
    for(i = 1; i <= n; i++)
        scanf("%d",&a[i]);
    int g = gcd(a[1],a[2]);
    for(i = 3; i <= n; i++)
        g = gcd(g,a[i]);//求出所有数的最大公因数
    for(i = 1; i <= n; i++)
        a[i] /= g;//将所有数除以最大公因数，对剩下的进行操作
    int flag = 1;
    for(i = 1; i <= n; i++){
        int t = a[i];
        while(t>1&&t%2==0)
            t /= 2;
        while(t>1&&t%3==0)//不断除二除三，看最后是否能变成1
            t /= 3;
        if(t!=1){//如果最后不是1说明这个数不是只由23构成，所以不能通过乘2或者3使他们相等
            flag = 0;
            break;
        }
    }
    if(flag)printf("Yes\n");
    else printf("No\n");
    return 0;
}