Edgy Trees

Description:
You are given a tree (a connected undirected graph without cycles) of n vertices. Each of the n−1 edges of the tree is colored in either black or red.

You are also given an integer k. Consider sequences of k vertices. Let’s call a sequence [a1,a2,…,ak] good if it satisfies the following criterion:

1. We will walk a path (possibly visiting same edge/vertex multiple times) on the tree, starting from a1 and ending at ak.
2. Start at a1, then go to a2 using the shortest path between a1 and a2, then go to a3 in a similar way, and so on, until you travel the shortest path between ak−1 and ak.
3. If you walked over at least one black edge during this process, then the sequence is good.

Consider the tree on the picture. If k=3 then the following sequences are good: [1,4,7], [5,5,3] and [2,3,7]. The following sequences are not good: [1,4,6], [5,5,5], [3,7,3].

There are nk sequences of vertices, count how many of them are good. Since this number can be quite large, print it modulo 1e9+7.

Input
The first line contains two integers n and k (2≤n≤1e5, 2≤k≤100), the size of the tree and the length of the vertex sequence.

Each of the next n−1 lines contains three integers ui, vi and xi (1≤ui,vi≤n, xi∈{0,1}), where ui and vi denote the endpoints of the corresponding edge and xi is the color of this edge (0 denotes red edge and 1 denotes black edge).

Output
Print the number of good sequences modulo 109+7.

Examples
input
4 4
1 2 1
2 3 1
3 4 1
output
252
input
4 6
1 2 0
1 3 0
1 4 0
output
0
input
3 5
1 2 1
2 3 0
output
210
Note
In the first example, all sequences (4^4) of length 4 except the following are good:

• [1,1,1,1]
• [2,2,2,2]
• [3,3,3,3]
• [4,4,4,4]

In the second example, all edges are red, hence there aren’t any good sequences.

Problem solving:

Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll    maxn = 1e5 + 10;
const ll    mod = 1e9 + 7;
ll          k, n, f[maxn];
map<ll, ll> ma;
ll find(ll x)
{
return f[x] == x ? x : f[x] = find(f[x]);
}
void join(ll x, ll y)
{
x    = find(x), y = find(y);
f[x] = y; ma[y] += ma[x];   // ma用来存储以y为根节点的连通块中节点的个数
}
ll poww(ll x, ll y)
{
ll ans = 1;
while (y)
{
if (y & 1)
ans = (ans * x) % mod;
x   = (x * x) % mod;
y >>= 1;
}
return ans;
}
int main()
{
ios::sync_with_stdio(0);
cin >> n >> k;
for (ll i = 1; i <= n; i++)
f[i] = i, ma[i] = 1;
ll ans = poww(n, k);
for (ll i = 1, u, v, w; i < n; i++)
{
cin >> u >> v >> w;
if (!w)
join(u, v);
}
for (ll i = 1; i <= n; i++)
if (f[i] == i)
ans = (ans - poww(ma[i], k) + mod) % mod;

cout << ans << endl;
return 0;
}