Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 1) A. Bear and Poker gcd

A. Bear and Poker

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai 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, ..., an (1 ≤ ai ≤ 109) — the bids of players.

Output

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

Sample test(s)

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)).复杂度为o(n);

#define N 100005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n;
ll  p[N],g;
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(S(n)!=EOF)
    {
        FI(n){
            scanf("%lld",&p[i]);
        }
        g= p[0];
        For(i,1,n){
            g = gcd(g,p[i]);
        }
        For(i,0,n){
            p[i] /= g;
        }
        bool flag = true;
        For(i,0,n){
            ll t = p[i];
            while(t > 1 && t % 2 == 0){
                t /= 2;
            }
            while(t > 1 && t % 3 == 0){
                t /= 3;
            }
            if(t != 1){
                flag = false;
                break;
            }
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}