cf615D Multipliers

Ayrat has number n, represented as it's prime factorization p_i of size m, i.e. n = p₁·p₂·...·p_m. Ayrat got secret information that that the product of all divisors of n taken modulo 10⁹ + 7 is the password to the secret data base. Now he wants to calculate this value.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of primes in factorization of n.

The second line contains m primes numbers p_i (2 ≤ p_i ≤ 200 000).

Output

Print one integer — the product of all divisors of n modulo 10⁹ + 7.

Example

Input

2
2 3

Output

36

Input

3
2 3 2

Output

1728

Note

In the first sample n = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.

In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.

P是n个质数的乘积，问P的所有因子之积是多少

先把质数整理下，假设质数p[i]出现rep[i]次

P的所有因子个数应当是∏(rep[i]+1)，记为S

然后对于一个质数p[i]，出现0个，1个，...rep[i]个p[i]的因子个数都是S/(rep[i]+1)

因此p[i]对于答案的贡献就是j=0~rep[i]∏(p[i]^j)^(S/(rep[i]+1))

= p[i]^(rep[i]*(rep[i]+1)/2*S/(rep[i]+1))

=p[i]^(rep[i]*S/2)

此时rep[i]*S/2太大，可能爆long long,所以还要处理：

根据欧拉定理，有a^phi(p)==1(mod p)，所以p[i]^(1e9+6)==1(mod 1e9+7)

所以S*rep[i]/2可以对1e9+6取模

但是1e9+6不是质数，除二不好做，所以对它的两倍2e9+12取模，防止除2之后丢失信息

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 #define mod 1000000007
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n,cnt;
 LL a[];
 LL p[],rep[];
 LL ans=;
 inline LL quickpow(LL a,LL b,LL MOD)
 {
     LL s=;
     a%=MOD;
     b=b%(MOD-);
     while (b)
     {
         if (b&)s=(s*a)%MOD;
         a=(a*a)%MOD;
         b>>=;
     }
     return s;
 }
 int main()
 {
     n=read();for (int i=;i<=n;i++)a[i]=read();
     sort(a+,a+n+);
     for (int i=;i<=n;i++)
     if (i==||a[i]!=a[i-])
     {
         p[++cnt]=a[i];
         rep[cnt]=;
     }else rep[cnt]++;
     LL pro=;
     for (int i=;i<=cnt;i++)pro=(pro*(rep[i]+))%(*mod-);
     for (int i=;i<=cnt;i++)
     {
         ans=ans*quickpow(p[i],pro*rep[i]/%(*mod-),mod)%mod;

     }
     printf("%lld\n",ans%mod);
 }

cf615D

也可以不把S/(rep[i]+1)和rep[i]+1约掉，搞一个{rep[i]+1}的前缀积、后缀积，就可以绕过除法把rep[i]+1挖掉

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<deque>
 #include<set>
 #include<map>
 #include<ctime>
 #define LL long long
 #define inf 0x7ffffff
 #define pa pair<int,int>
 #define mkp(a,b) make_pair(a,b)
 #define pi 3.1415926535897932384626433832795028841971
 #define mod 1000000007
 using namespace std;
 inline LL read()
 {
     LL x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL n,cnt;
 LL a[];
 LL p[],rep[];
 LL s[],t[];
 LL phimod=;
 LL mod2=;
 LL ans=;
 inline LL quickpow(LL a,LL b,LL MOD)
 {
     LL s=;
     a%=MOD;
     b=b%(MOD-);
     while (b)
     {
         if (b&)s=(s*a)%MOD;
         a=(a*a)%MOD;
         b>>=;
     }
     return s;
 }
 int main()
 {
     n=read();for (int i=;i<=n;i++)a[i]=read();
     sort(a+,a+n+);
     for (int i=;i<=n;i++)
     if (i==||a[i]!=a[i-])
     {
         p[++cnt]=a[i];
         rep[cnt]=;
     }else rep[cnt]++;
     s[]=t[cnt+]=;
     for (int i=;i<=cnt;i++)
     {
         s[i]=(s[i-]*(rep[i]+))%(mod-);
     }
     for (int i=cnt;i>=;i--)
         t[i]=t[i+]*(rep[i]+)%(mod-);
     for (int i=;i<=cnt;i++)
     {
         LL ap=s[i-]*t[i+]%(mod-);
         ans=ans*quickpow(p[i],(rep[i]+)*rep[i]/%(mod-)*ap,mod)%mod;
     }
     printf("%lld\n",ans%mod);
 }

cf615D_2