Codeforces Round # (Div. 2) D. Remainders Game (Chinese remainder theorem)

Source: Internet
Author: User


D. remainders Game


Today Pari and Arya are playing a game called remainders.



Pari chooses, positive integer  x  and  K , and tells Arya  k  but not  x . Arya has to find the value . There Are  n  ancient numbers  c Span class= "Lower-index" >1, C 2, ..., c N  and Pari have to tell Arya  if Arya wants. Given  K  and The ancient values, tell us if Arya have a winning strategy Independe NT of value Of  x  or not. Formally, is it true this Arya can understand the value  for any positive integer  x ?



Note, that's means the remainder of x after dividing it by y.


Input


The first line of the input contains the integers n and k (1≤ n, K ≤1 -the number of ancient integers and value k that's chosen by Pari.



The second line contains n integers c1, C2, ..., c n (1≤ ci ≤1).


Output


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


Examplesinput
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. For example 1 and 7 has the same remainders after dividing by 2 and 3, but they differ in remainders afte R dividing by 7.


#include <cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>#include<iostream>using namespacestd;#definell Long Longll LCM;Const intMAXN = 1e6 +Ten; ll GETLCM (ll A, ll b) {returnA/__GCD (A, b) *b;} ll P[MAXN];intMain () {ll n, K;  while(~SCANF ("%i64d%i64d", &n, &k)) { for(inti =1; I <= N; i++) scanf ("%i64d", &P[i]); ll LCM=1; intFlag =0;  for(inti =1; I <= N; i++) {LCM=GETLCM (LCM, p[i]); LCM= LCM%K; if(LCM = =0) {puts ("Yes"); Flag=1;  Break; }        } if(!flag) puts ("No"); }}





Codeforces Round #(Div. 2) D. Remainders Game (Chinese remainder theorem)


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