Problem:
Score : 900 points

Problem Statement

We have an undirected weighted graph with N vertices and M edges.
The i-th edge in the graph connects Vertex U_i and Vertex V_i, and has a weight of W_i.
Additionally, you are given an integer X.

Find the number of ways to paint each edge in this graph either white or black such that the following condition is met, modulo 10^9 + 7:

The graph has a spanning tree that contains both an edge painted white and an edge painted black. Furthermore, among such spanning trees, the one with the smallest weight has a weight of X.

Here, the weight of a spanning tree is the sum of the weights of the edges contained in the spanning tree.

Constraints

1 \leq N \leq 1 000

1 \leq M \leq 2 000

1 \leq U_i, V_i \leq N (1 \leq i \leq M)

1 \leq W_i \leq 10^9 (1 \leq i \leq M)

If i \neq j, then (U_i, V_i) \neq (U_j, V_j) and (U_i, V_i) \neq (V_j, U_j).

U_i \neq V_i (1 \leq i \leq M)

The given graph is connected.

1 \leq X \leq 10^{12}

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
X
U_1 V_1 W_1
U_2 V_2 W_2
:
U_M V_M W_M

Output

Print the answer.

Sample Input 1

3 3
2
1 2 1
2 3 1
3 1 1

Sample Output 1

6

Sample Input 2

3 3
3
1 2 1
2 3 1
3 1 2

Sample Output 2

2

Sample Input 3

5 4
1
1 2 3
1 3 3
2 4 6
2 5 8

Sample Output 3

0

Sample Input 4

8 10
49
4 6 10
8 4 11
5 8 9
1 8 10
3 8 128773450
7 8 10
4 2 4
3 4 1
3 1 13
5 2 2

Sample Output 4

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
int n,m;
struct edge
{
	int u,v,w;
	bool usd;
	edge(){}
	edge(int u,int v,int w):u(u),v(v),w(w){}
	bool operator <(const edge&t)const{return w<t.w;}
}es[2200];
int fa[1100];
vector<pii > g[1100];
ll X,val[1100];
int pa[12][1100],wei[12][1100],dep[1100];
void dfs(int x,int fr,int fw)
{
	pa[0][x]=fr;
	wei[0][x]=fw;
	dep[x]=dep[fr]+1;
	for(int i=0;i<g[x].size();i++)
	{
		int to=g[x][i].first;
		if(to==fr)continue;
		dfs(to,x,g[x][i].second);
	}
}
inline int getfa(int x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
pii getmax(int x,int y)
{
	if(dep[x]<dep[y])swap(x,y);
	int ret=0;
	for(int i=0;i<12;i++)
	{
		if((dep[x]-dep[y])>>i&1)
		{
			ret=max(ret,wei[i][x]);
			x=pa[i][x];
		}
	}
	if(x==y)return make_pair(ret,x);
	for(int k=11;k>=0;k--)
	{
		if(pa[k][x]!=pa[k][y])
		{
			ret=max(ret,wei[k][x]);
			ret=max(ret,wei[k][y]);
			x=pa[k][x];
			y=pa[k][y];
		}
	}
	ret=max(ret,wei[0][x]);
	ret=max(ret,wei[0][y]);
	return make_pair(ret,pa[0][x]);
}
int rht[2200];
inline void add(int &a,int b)
{
	a+=b;
	if(a>=MOD)a-=MOD;
}
inline int quickpow(ll base,ll ind)
{
	ll ans=1;
	while(ind>0)
	{
		if(ind&1)ans=ans*base%MOD;
		ind>>=1;
		base=base*base%MOD;
	}
	return (int)ans;
}
vector<int> repl[2020];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%lld",&X);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d%d",&es[i].u,&es[i].v,&es[i].w);
	}
	sort(es,es+m);
	rht[m-1]=m;
	for(int i=m-2;i>=0;i--)
	{
		if(es[i].w==es[i+1].w)rht[i]=rht[i+1];
		else rht[i]=i+1;
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	int ru,rv;
	ll sum=0;
	for(int i=0;i<m;i++)
	{
		if((ru=getfa(es[i].u))!=(rv=getfa(es[i].v)))
		{
			fa[ru]=rv;
			es[i].usd=1;
			sum+=es[i].w;
			g[es[i].u].push_back(make_pair(es[i].v,es[i].w));
			g[es[i].v].push_back(make_pair(es[i].u,es[i].w));
		}
		else
		{
			es[i].usd=0;
		}
	}
	dfs(1,0,0);
	for(int k=1;k<12;k++)
	{
		for(int i=1;i<=n;i++)
		{
			pa[k][i]=pa[k-1][pa[k-1][i]];
			wei[k][i]=max(wei[k-1][i],wei[k-1][pa[k-1][i]]);
		}
	}
	int ans=0;
	for(int i=0;i<m;i++)
	{
		if(es[i].usd)
		{
			val[i]=sum;
		}
		else
		{
			pii tmp=getmax(es[i].u,es[i].v);
			val[i]=sum+es[i].w-tmp.first;
			int x=es[i].u;
			while(x!=tmp.second)
			{
				if(wei[0][x]==tmp.first)repl[i].push_back(x);
				x=pa[0][x];
			}
			x=es[i].v;
			while(x!=tmp.second)
			{
				if(wei[0][x]==tmp.first)repl[i].push_back(x);
				x=pa[0][x];
			}
		}
	//	cout<<i<<" "<<val[i]<<endl;
	}
	for(int i=0;i<m;i++)
	{
		if(val[i]==X)
		{
			int res=2;
			for(int j=0;j<m;j++)
			{
				if(val[j]>X||(val[j]==X&&j>i))
					res=res*2%MOD;
			}
			add(ans,res);
		}
	}
	printf("%d\n",ans);
	return 0;
}