簡單介紹二項式反演： 如果存在$a_n = \sum_{i = s}^n\left(\begin{array}{l} n \\ i \end{array}\right)b_i$則可以得到$b_n = \sum_{i = s} ^ n(-1)^{n - i}\left(\begin{array}{l} n \\ i \end{array}\right)a_i$

題意：給出一個含有n個節點的樹，以及k個顏色，詢問有多少種方式正好用k個顏色給樹染色，並且任意兩個相鄰的節點顏色不同。

題解：我們發現只要一個節點與他的父節點顏色不同即可，所以對於根節點有$k*(k - 1)^{n - 1}$種方案，我們假設$f(n)$為恰好用n種顏色染色整棵樹的方案數，$g(n)$為至多用n種顏色染色整棵樹的方案數，顯然$g(n) = \sum_{i = 2}^{n}\left(\begin{array}{l} n \\ i \end{array}\right)f(i)$由二項式反演可以得到$f(n) = \sum_{i = 2} ^ n(-1)^{n - i}\left(\begin{array}{l} n \\ i \end{array}\right)g(i)$

$ac code:$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define met(a, b) memset(a, b, sizeof(a))
 

#define rep(i, a, b) for(int i = a; i <= b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
#define fi first
#define se second
#define pb push_back
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

ll qp(ll base, ll n) {ll res = 1; while(n) {if(n & 1) res = (res * base) % mod; base = (base * base) % mod; n >>= 1;} return res;};
ll inv(ll x) {return qp(x, mod - 2);}
ll fac(ll n) {ll res = 1; for(int i = 1; i <= n; i++) res = (res * i) % mod; return res;}
ll C(ll n, ll m){return fac(n) * inv(fac(m)) % mod * inv(fac(n - m)) % mod;}

vector<int> G[maxn];
ll dp[maxn];

ll dfs(int u, int col) {
    dp[u] = 1;
    for(auto v : G[u]) {
        dfs(v, col);
        dp[u] = dp[u] * dp[v] % mod * max(col - 1, 0) % mod;
    }
    return dp[u] * col % mod;
}

int main() {
    int n, k, p;
    while(~scanf("%d%d", &n, &k)) {
        rep(i, 1, maxn - 1) G[i].clear();
        rep(i, 1, n - 1) {
            scanf("%d", &p);
            G[p].pb(i);
        }
        ll sum = 0, f = 1;
        rep(i, 0, k) {
            f = (i % 2) ? -1 : 1;
            sum = (sum + f * C(k, i) % mod * dfs(0, k - i) % mod + mod) % mod;
        }
        printf("%lld\n", sum);
    }
    return 0;
}