F. Cowmpany Cowmpensation dp+拉格朗日插值

题意:一个数,每个节点取值是1-d,父亲比儿子节点值要大,求方案数
题解:(dp[u][x]=prod_{v}sum_{i=1}^xdp[v][i]),v是u的子节点,先预处理出前3000项,如果d大于3000,用这些结果插值即可

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;
//using namespace __gnu_pbds;

const ull ba=233;
const db eps=1e-5;
const ld pi=acos(-1);
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=3000+10,maxn=3000+10,inf=0x3f3f3f3f;

vi v[N];
ll dp[N][N],l[N],inv[N*2];
void dfs(int u)
{
    for(int i=1;i<N;i++)dp[u][i]=1;
    for(int x:v[u])
    {
        dfs(x);
        for(int i=1;i<N;i++)dp[u][i]=dp[u][i]*dp[x][i]%mod;
    }
    for(int i=1;i<N;i++)add(dp[u][i],dp[u][i-1]);
}
int main()
{
    for(int i=-N;i<N;i++)inv[i+N]=qp(i,mod-2);
    int n,d;scanf("%d%d",&n,&d);
    for(int i=2;i<=n;i++)
    {
        int x;scanf("%d",&x);
        v[x].pb(i);
    }
    dfs(1);
    if(d<N)printf("%lld
",dp[1][d]);
    else
    {
        ll ans=0;
        for(int i=1;i<N;i++)
        {
            l[i]=1;
            for(int j=1;j<N;j++)if(i!=j)
                l[i]=l[i]*(d-j)%mod*inv[i-j+N]%mod;
            l[i]=(l[i]+mod)%mod;
            add(ans,dp[1][i]*l[i]%mod);
        }
        printf("%lld
",ans);
    }
    return 0;
}
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