一次操作对答案的贡献是
[prodlimits_{i=1,i
e x}^{n}a_i=prodlimits_{i=1}^{n}a_i-(a_x-1)prodlimits_{i=1,i
e x}^{n}a_i
]
我们设等式右侧的减号左边为(S_1),右侧为(S_2)会得到总贡献为
[S_1-S_2+S_2-S_3+S_3-S_4+……+S_k-S_{k+1}
]
假设第(i)个数被选择(b_i)次，那么答案就是
[prod a_i-prod(a_i-b_i)
]
我们只要求后面那部分的期望
[E=frac{1}{n^k}sumlimits_{sum b_i=k}frac{k!}{prod (b_i)!}prod (a_i-b_i)
]
[=frac{k!}{n^k}sumlimits_{sum b_i=k}prod frac{(a_i-b_i)}{b_i}
]
将他写成指数生成函数的形式
[=frac{k!}{n^k}[x^k]prodlimits_{i=1}^{n}sumlimits_{j=0}frac{a_i-j}{j!}x^j
]
[=frac{k!}{n^k}[x^k]prodlimits_{i=1}^{n}sumlimits_{j=0}(a_ifrac{x^j}{j!}-jfrac{x^j}{j!})
]
[=frac{k!}{n^k}[x^k]prodlimits_{i=1}^{n}(a_ie^x-sumlimits_{j=0}frac{x^j}{(j-1)!})
]
[=frac{k!}{n^k}[x^k]prodlimits_{i=1}^{n}(a_ie^x-sumlimits_{j=0}xfrac{x^j}{j!})
]
[=frac{k!}{n^k}[x^k]prodlimits_{i=1}^{n}(a_ie^x-xe^x)
]
[=frac{k!}{n^k}[x^k](e^{nx}prodlimits_{i=1}^{n}(a_i-x))
]
右边的(prodlimits_{i=1}^{n}(a_i-x))可以(O(n^2))乘出来，设(f(x))为得到的多项式
[=frac{k!}{n^k}[x^k](sumlimits_{i=0}frac{(nx)^i}{i!}sumlimits_{j=0}f_jx^j)
]
[=frac{k!}{n^k}[x^k](sumlimits_{t=0}x^tsumlimits_{i+j=t}frac{n^if_j}{i!})
]
[=sumlimits_{i+j=k}frac{k^{underline{k-i}}c_j}{n^{k-i}}
]
[=sumlimits_{i=0}^{min(n,k)}frac{k^{underline{i}}c_i}{n^i}
]
所以
[ans=prod a_i-sumlimits_{i=0}^{min(n,k)}frac{k^{underline{i}}c_i}{n^i}
]
#include<bits/stdc++.h>
using namespace std;
namespace red{
#define int long long
#define ls(p) (p<<1)
#define rs(p) (p<<1|1)
#define lowbit(i) ((i)&(-i))
inline int read()
{
int x=0;char ch,f=1;
for(ch=getchar();(ch<'0'||ch>'9')&&ch!='-';ch=getchar());
if(ch=='-') f=0,ch=getchar();
while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f?x:-x;
}
const int N=4e5+10,p=1e9+7;
int n,k,ans,len;
int fac[N],inv[N];
int a[N];
int f[N],g[N],h[N];
inline void add(int &x,int y)
{
x+=y;
if(x>=p) x-=p;
if(x<0) x+=p;
}
inline int fast(int x,int k)
{
int ret=1;
while(k)
{
if(k&1) ret=ret*x%p;
x=x*x%p;
k>>=1;
}
return ret;
}
inline void main()
{
n=read(),k=read();
ans=1;f[0]=1;len=0;
for(int i=1;i<=n;++i)
{
a[i]=read();
ans=ans*a[i]%p;
g[0]=a[i],g[1]=-1;
for(int j=0;j<i;++j)
{
for(int k=0;k<=1;++k)
{
add(h[j+k],f[j]*g[k]%p);
}
}
for(int j=0;j<=i;++j) f[j]=h[j],h[j]=0;
}
for(int tmp=1,i=0;i<=min(n,k);++i,tmp=tmp*n%p)
{
int ret=f[i];
for(int j=k,d=i;d;--d,--j) ret=ret*j%p;
ret=ret*fast(tmp,p-2);
add(ans,-ret%p);
}
printf("%lld
",ans);
}
}
signed main()
{
red::main();
return 0;
}