Codeforces 487C 数论+构造

Prefix Product Sequence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a sequence [a1, a2, ... , an]. Define its prefix product sequence .

Now given n, find a permutation of [1, 2, ..., n], such that its prefix product sequence is a permutation of [0, 1, ..., n - 1].

Input

The only input line contains an integer n (1 ≤ n ≤ 105).

Output

In the first output line, print "YES" if such sequence exists, or print "NO" if no such sequence exists.

If any solution exists, you should output n more lines. i-th line contains only an integer ai. The elements of the sequence should be different positive integers no larger than n.

If there are multiple solutions, you are allowed to print any of them.

Examples

input

7

output

YES1436527

input

6

output

NO

Note

For the second sample, there are no valid sequences.

题意：给你一个n，构造出1-n的排列使得a1%n  a1a2%n .... a1a2...an%n 生成的序列是0-(n-1)的一个排列

题解：特判1和4  如果n是合数就为NO，因为不管怎么排列前面项乘起来都会变成n的倍数，造成出现多个0,

因此n这个数也必须放到最后面，即an=n。

如果n是素数

令 ai=i/(i-1)(mod n)

i乘上i-1的逆元即可

a1*a2*a3*....ai=1*inv(1)*2*inv(2)....*i-1*inv(i-1)*i=i

所以此序列也是合法序列

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;ll judge(ll n){for(ll i=2;i*i<=n;i++){if(n%i==0)return 0;}return 1;}ll quick(ll a,ll k,ll mod){ll ans=1;while(k){if(k&1)ans=ans*a%mod;a=a*a%mod;k/=2;}return ans;}int main(){ll n;scanf("%lld",&n);if(n==4){printf("YES\n1\n3\n2\n4\n");return 0;}else if(n==1){printf("YES\n1\n");return 0;}if(judge(n)){ll now=1;printf("YES\n1\n");for(ll i=2;i<n;i++){now=quick(i-1,n-2,n);printf("%lld\n",i*now%n);}printf("%lld\n",n);}else{printf("NO\n");}return 0;}