Nefu 120 Mason Prime number (--lucas-lehmer test judgment)

Mason Prime

problem:120 Time limit:1000ms Memory limit:65536k

Description

As Mason learned, talented, enthusiastic and the first systematic and deep study of the number of 2p-1 (of which p is a prime number), in order to commemorate him, the number of scholars called this number "Mason numbers", and the MP (which M is the first letter of the Mason name), namely Mp=2p-1. If Mason number is prime, it is called "Mason Prime". For example p=2,3,5,7, MP is prime, but 211-1 is not prime. Now ask you to find the first n of the number of Mason primes.

Input

There are multiple sets of test data. The first line is a positive integer t, which represents the number of groups of test data. Next to each set of 1 number p values, here 2<= p <= 62.

Output

For each set of test data, to determine whether the MP is the number of Mason, is the output "yes", no output "No", the output to be changed line.

Sample_input

227

Sample_output

Yesyes

Idea: Obviously n=2^p-1 when P is not a prime number, N is not a Mason, obviously not a prime (can decompose the proof of the type)

When P is prime, n is the number of Mason, it is necessary to determine whether n is prime, which can be determined by Lucas-lehmer test.

Knowledge Involved:

The Lucas-xylem test principle is this: make the Mason number m_{p = 2^{p1 as the test object (the preset p is the prime, otherwise Mp is the composite). Define sequence {si }: All i ≥0}}

if; , if.

.

.

.

The first few items of this sequence are 4, 14, 194, 37634, ... (Sequence A003010 in OEIS) then M_{p is prime number and only if}

Otherwise, M_{p is composite. s P . 2 the remainder of modulo Mp is called the Lucas-xylem remainder of p .}

Use the program to calculate the Sp-2 can, to pay attention to is p=2, can not use this method of determination, a special sentence can be

Accepted 800k 4ms C + + (g++ 3.4.3) 692 #include <cstdio> #include <iostream> #include <cstring> #include <algorithm>using namespace Std;typedef Long long ll;ll t,p;ll multi (ll a,ll b,ll MoD)//handwriting multiplication calculation, because the direct multiply will overflow long long{
   ll ans=0;    while (b)    {        if (b&1)            ans= (ans+a)%mod;        b>>=1;        A= (a<<1)%mod;    }    return ans;} int main () {    scanf ("%lld", &t);    while (t--)    {        scanf ("%lld", &p);        ll n= ((LL) 1<<p)-1;        ll r=4;        if (p==2) puts ("Yes"), and/or the other        {while            ((--p)!=1) r= (multi (r,r,n) -2+n)%n;            if (r%n==0) puts ("yes");            Else puts ("no");        }    }    return 0;}

Nefu 120 Mason Prime number (--lucas-lehmer test judgment)