[codeforces1139C]Edgy Trees

time limit per test : 2 seconds
memory limit per test : 256 megabytes

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

You are also given an integer kkk . Consider sequences of kkk vertices. Let’s call a sequence [a1,a2,…,ak][a_1,a_2,…,a_k][a1,a2,…,ak]
good if it satisfies the following criterion:

We will walk a path (possibly visiting same edge/vertex multiple times) on the tree, starting from a1a_1a1 and ending at aka_kak.
Start at a1a_1a1, then go to a2a_2a2 using the shortest path between a1a_1a1 and a2a_2a2, then go to a3a_3a3 in a similar way, and so on, until you travel the shortest path between ak−1a_{k−1}ak−1 and aka_kak.
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=3k=3k=3 then the following sequences are good: [1,4,7],[5,5,3][1,4,7], [5,5,3][1,4,7],[5,5,3] and [2,3,7][2,3,7][2,3,7]. The following sequences are not good: [1,4,6],[5,5,5],[3,7,3][1,4,6], [5,5,5], [3,7,3][1,4,6],[5,5,5],[3,7,3].

There are nkn^knk
sequences of vertices, count how many of them are good. Since this number can be quite large, print it modulo 109+710^9+7109+7.

Input

The first line contains two integers nnn and k(2≤n≤105,2≤k≤100)k (2≤n≤10^5, 2≤k≤100)k(2≤n≤105,2≤k≤100), the size of the tree and the length of the vertex sequence.
Each of the next n−1n−1n−1 lines contains three integers uiu_iui, viv_ivi and xi(1≤ui,vi≤n,xi∈0,1)x_i (1≤u_i,v_i≤n, x_i∈{0,1})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 (000 denotes red edge and 111 denotes black edge).
Output

Print the number of good sequences modulo 109+70^9+709+7.

Examples

Output

Input

Output

Input

3 5
1 2 1
2 3 0

Output

Note

In the first example, all sequences (444^444) of length 444
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.

题意：
给定一个n个节点的树，有n-1个染色的无向边（红色或者黑色），询问有多少个长度为k的序列a1,a2,a3...aka_1,a_2,a_3...a_ka1,a2,a3...ak（aia_iai走到ai+1a_{i+1}ai+1的路径是他们之间的最短路。）序列中元素可以重复，使得按顺序走过这个序列里面所有的点的路径中至少有一条黑色边。
题解：
并查集。可以知道如果不考虑其他东西的话一共有nkn^knk种路径，先把所有的红色边加入并查集，那么每个联通块内的元素所构成的序列一定是没有黑色边的。统计每个联通块内能组成的长度为kkk的不同的路径有多少条。然后用总条数减去即可。（记得要即时取模…不然会炸。）

#include<bits/stdc++.h>
#define LiangJiaJun main
#define ll long long
#define MOD 1000000007LL
using namespace std;
int n,k;
struct edge{int u,v,w;
}e[200004];
int sz[100004];
ll ans,sum;
int f[100004];
int Find(int x){return f[x]==x?f[x]:f[x]=Find(f[x]);}
void Union(int x,int y){int p=Find(x),q=Find(y);if(p!=q){f[p]=q;sz[q]+=sz[p];sz[p]=0;}
}
ll  fp(ll x,ll y){if(y==0)return 1;ll temp=fp(x,y>>1);if(~y&1)return (temp*temp)%MOD;else return (((temp*temp)%MOD)*x)%MOD;
}
int w33ha(){ans=0;sum=0;for(int i=1;i<=n;i++){f[i]=i;sz[i]=1;}for(int i=1;i<n;i++){scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);if(e[i].w==0){Union(e[i].u,e[i].v);}}for(int i=1;i<=n;i++){if(sz[i]!=0){sum=(sum+fp(1LL*sz[i],k))%MOD;}}ans=1;for(int i=1;i<=k;i++){ans=(ans*n)%MOD;}ans=(ans-sum+MOD)%MOD;printf("%lld\n",ans);return 0;
}
int LiangJiaJun(){while(scanf("%d%d",&n,&k)!=EOF)w33ha();return 0;
}

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