CodeForces

Discription

You are given an undirected graph, constisting of n vertices and m edges. Each edge of the graph has some non-negative integer written on it.

Let's call a triple (u, v, s) interesting, if 1 ≤ u < v ≤ n and there is a path (possibly non-simple, i.e. it can visit the same vertices and edges multiple times) between vertices u and v such that xor of all numbers written on the edges of this path is equal to s. When we compute the value s for some path, each edge is counted in xor as many times, as it appear on this path. It's not hard to prove that there are finite number of such triples.

Calculate the sum over modulo 10⁹ + 7 of the values of s over all interesting triples.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 0 ≤ m ≤ 200 000) — numbers of vertices and edges in the given graph.

The follow m lines contain three integers u_i, v_i and t_i (1 ≤ u_i, v_i ≤ n, 0 ≤ t_i ≤ 10¹⁸, u_i ≠ v_i) — vertices connected by the edge and integer written on it. It is guaranteed that graph doesn't contain self-loops and multiple edges.

Output

Print the single integer, equal to the described sum over modulo 10⁹ + 7.

Examples

Input

4 4
1 2 1
1 3 2
2 3 3
3 4 1

Output

12

Input

4 4
1 2 1
2 3 2
3 4 4
4 1 8

Output

90

Input

8 6
1 2 2
2 3 1
2 4 4
4 5 5
4 6 3
7 8 5

Output

62

Note

In the first example the are 6 interesting triples:

1. (1, 2, 1)
2. (1, 3, 2)
3. (1, 4, 3)
4. (2, 3, 3)
5. (2, 4, 2)
6. (3, 4, 1)

The sum is equal to 1 + 2 + 3 + 3 + 2 + 1 = 12.

In the second example the are 12 interesting triples:

1. (1, 2, 1)
2. (2, 3, 2)
3. (1, 3, 3)
4. (3, 4, 4)
5. (2, 4, 6)
6. (1, 4, 7)
7. (1, 4, 8)
8. (2, 4, 9)
9. (3, 4, 11)
10. (1, 3, 12)
11. (2, 3, 13)
12. (1, 2, 14)

The sum is equal to 1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 + 11 + 12 + 13 + 14 = 90.

 

 

    根据 Wc 2012 Xor 那道题，我们知道，两点之间所有路径xor和是可以通过随意找一条路径然后放到环的线性基里xjb异或而得到的，这个不难画图发现(模拟中途走去环然后返回的过程)。

 

    于是，本题就是对于每个联通分量，先随便搞出一颗dfs树，并把所有回边构成的环加到线性基里。

    然后发现任意两点之间在dfs树上路径的xor和就等于他们到根的xor和再xor起来，之后再考虑上线性基，推完一波式子之后直接做就行了。

 

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=100005,ha=1e9+7;
int hd[maxn],to[maxn*4],ne[maxn*4],num,cnt[69][2],tot,n,m,ans;
ll val[maxn*4],ci[69],a[69],Xor[maxn];
bool v[maxn],can[69];

inline int add(int x,int y){ x+=y; return x>=ha?x-ha:x;}
inline void ADD(int &x,int y){ x+=y; if(x>=ha) x-=ha;}
inline void addline(int x,int y,ll z){
    to[++num]=y,ne[num]=hd[x],hd[x]=num,val[num]=z;
}

inline int C(int x){ return (x*(ll)(x-1)>>1)%ha;}

inline void update(ll x){
	for(int i=0;i<=60;i++) if(x&ci[i]) can[i]=1;
}

inline void ins(ll x){
	for(int i=60;i>=0;i--) if(x&ci[i]){
		if(!a[i]){ a[i]=x,tot++,update(x); return;}
		x^=a[i];
	}
}

void dfs(int x,int fa){
	v[x]=1;
	for(int i=hd[x];i;i=ne[i]) if(to[i]!=fa){
		if(!v[to[i]]) Xor[to[i]]=Xor[x]^val[i],dfs(to[i],x);
		else ins(Xor[x]^Xor[to[i]]^val[i]);
	}
	
	for(int i=0;i<=60;i++) cnt[i][(Xor[x]&ci[i])?1:0]++;
}

inline void solve(){
	for(int o=1;o<=n;o++) if(!v[o]){
        memset(cnt,0,sizeof(cnt));
        memset(a,0,sizeof(a)),tot=0;
        memset(can,0,sizeof(can));
		dfs(o,0);
        
        /*
		for(int i=0;i<=60;i++)
	        for(int j=1;j<=n;j++) cnt[i][(Xor[j]&ci[i])?1:0]++;
	    */
	    
    	for(int i=0,now;i<=60;i++){
            now=0;
            
            if(can[i]){
            	ADD(now,cnt[i][0]%ha*(ll)cnt[i][1]%ha);
            	ADD(now,add(C(cnt[i][0]),C(cnt[i][1]))%ha);
            	now=now*(ll)(ci[tot-1]%ha)%ha;
    		}
    		else ADD(now,ci[tot]%ha*(ll)cnt[i][0]%ha*(ll)cnt[i][1]%ha);
    		
    		ADD(ans,now*(ll)(ci[i]%ha)%ha);
    	}
    }
}

int main(){
	ci[0]=1;
	for(int i=1;i<=60;i++) ci[i]=ci[i-1]+ci[i-1];
	
	scanf("%d%d",&n,&m);
	int uu,vv; ll ww;
	for(int i=1;i<=m;i++){
	    scanf("%d%d%I64d",&uu,&vv,&ww);
		addline(uu,vv,ww),addline(vv,uu,ww);
    }
    
	solve();
	
	printf("%d
",ans);
	return 0;
}