$我们要< x , y >路径上的点全部同色即根据DFS序来涂即可$
$f[i][j]:前i个点用了j个颜色$
$转移：f[i][j] = f[i-1][j] + f[i-1][j-1] * (k - j + 1);$
图片来自牛客本题讨论

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int mod = 1e9 + 7;
const int N = 310;

vector<int> g[N];
int n, k;
int f[N][N];

int main()
{
    cin >> n >> k;
    for(int i = 1; i <= n; i ++) 
    {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    f[0][0] = 1;

    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= k; j ++)
        {
            f[i][j] = (ll)(f[i-1][j] + (ll)f[i-1][j-1] * (k - j + 1) % mod)%mod;
        }

    ll ans = 0;

    for(int i = 1; i <= k; i ++) ans = (ans + f[n][i]) % mod;

    cout << ans % mod << endl;
}