Problem:
Score : 1200 points

Problem Statement

You are given a simple connected undirected graph G consisting of N vertices and M edges.
The vertices are numbered 1 to N, and the edges are numbered 1 to M.

Edge i connects Vertex a_i and b_i bidirectionally.
It is guaranteed that the subgraph consisting of Vertex 1,2,\ldots,N and Edge 1,2,\ldots,N-1 is a spanning tree of G.

An allocation of weights to the edges is called a good allocation when the tree consisting of Vertex 1,2,\ldots,N and Edge 1,2,\ldots,N-1 is a minimum spanning tree of G.

There are M! ways to allocate the edges distinct integer weights between 1 and M.
For each good allocation among those, find the total weight of the edges in the minimum spanning tree, and print the sum of those total weights modulo 10^{9}+7.

Constraints

All values in input are integers.

2 \leq N \leq 20

N-1 \leq M \leq N(N-1)/2

1 \leq a_i, b_i \leq N

G does not have self-loops or multiple edges.

The subgraph consisting of Vertex 1,2,\ldots,N and Edge 1,2,\ldots,N-1 is a spanning tree of G.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
\vdots
a_M b_M

Output

Print the answer.

Sample Input 1

3 3
1 2
2 3
1 3

Sample Output 1

6

An allocation is good only if Edge 3 has the weight 3. For these good allocations, the total weight of the edges in the minimum spanning tree is 3, and there are two good allocations, so the answer is 6.

Sample Input 2

4 4
1 2
3 2
3 4
1 3

Sample Output 2

50

Sample Input 3

15 28
10 7
5 9
2 13
2 14
6 1
5 12
2 10
3 9
10 15
11 12
12 6
2 12
12 8
4 10
15 3
13 14
1 15
15 12
4 14
1 7
5 11
7 13
9 10
2 7
1 9
5 6
12 14
5 2

Sample Output 3

657573092

Print the sum of those total weights modulo 10^{9}+7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC target("avx")  // CPU 処理並列化
#pragma GCC optimize("O3")  // CPU 処理並列化
#pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double INF=1e+10;
const long double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	inline long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
int L[110];
int R[110];
pair<long long,long long> dp[2][1<<19];
int K[1<<19];
int TMP[30];
vector<int>to[1<<19];
int main(){
	int a,b;
	scanf("%d%d",&a,&b);
	for(int i=0;i<b;i++){
		scanf("%d%d",L+i,R+i);
		L[i]--;R[i]--;
	}
	int n=a-1;
	for(int i=0;i<(1<<n);i++){
		init_UF(a);
		for(int j=0;j<n;j++){
			if(i&(1<<j))UNION(L[j],R[j]);
			else to[i].push_back(1<<j);
		}
		for(int j=0;j<a;j++)TMP[j]=FIND(j);
		for(int j=0;j<b;j++){
			if(TMP[L[j]]==TMP[R[j]])K[i]++;
		}
	}
 
	dp[0][0].first=1;
	for(int i=0;i<b;i++){
		int h=i%2;
		int g=!h;
		for(int j=(1<<n)-1;j>=0;j--){
			if(!dp[h][j].first){
				dp[g][j].first=dp[g][j].second=0;
				continue;
			}
			if(K[j]>i){
				dp[g][j].first=dp[h][j].first*(K[j]-i);
				dp[g][j].second=dp[h][j].second*(K[j]-i);
			}else{
				// dp[g][j]=make_pair(0LL,0LL);

			}
			long long ad=dp[h][j].second+dp[h][j].first*(i+1);
			for(int k=0;k<to[j].size();k++){
				dp[g][j|to[j][k]].first+=dp[h][j].first;
				dp[g][j|to[j][k]].second+=ad;
			}
		}
		for(int j=0;j<(1<<n);j++){
			dp[g][j].first%=mod;
			dp[g][j].second%=mod;
		}
	}
	printf("%lld\n",dp[b%2][(1<<n)-1].second%mod);
}