/*
考虑一下已经放回m本书的情况，已经有书的格子不要管他，考虑没有书的格子，不考虑错排有(n-m)!种，在逐步考虑有放回原来位置的情况，已经放出去和已经被占好的格子，不用考虑，剩下全都考虑，设t=x∩y，把除t以外的搞一下容斥就行了
*/
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define ll long long
#define fo(i,l,r) for(int i = l;i <= r;i++)
#define fd(i,l,r) for(int i = r;i >= l;i--)
using namespace std;
const int maxn = ;
const ll mod = 1000000007LL;
ll read(){
ll x=,f=;
char ch=getchar();
while(!(ch>=''&&ch<='')){if(ch=='-')f=-;ch=getchar();};
while(ch>=''&&ch<=''){x=x*+(ch-'');ch=getchar();};
return x*f;
}
ll n,m,x[maxn],y[maxn],rec[maxn],c[maxn];
bool usd[maxn];
ll ans,fac[maxn];
ll q_pow(ll a,ll b){
ll ret = ;
while(b){
if(b&) ret = (ret*a) % mod;
a = (a*a) % mod;
b >>= ;
}
return ret;
}
inline ll inv(ll x){
return q_pow(x,mod-);
}
int main(){
freopen("problem.in","r",stdin);
freopen("problem.out","w",stdout);
n = read();m = read();
fo(i,,m) x[i] =read(),y[i]=read();
fo(i,,m){
usd[x[i]] = true;
usd[y[i]] = true;
}
int t = n;
fo(i,,n) if(usd[i]) t--;
fac[] = fac[] = ;
fo(i,,n) fac[i] = (fac[i-]*i) % mod;
c[] = c[t] = ;
if(t > ) c[] = c[t-] = t;
fo(i,,t-)c[i] = ((c[i-]*(t-i+)%mod)*inv(i))%mod;
ans = fac[n-m];
fo(i,,t){
if(i&) ans = (ans + *mod - (c[i]*fac[n-m-i]) % mod) % mod;
else ans = (ans + (c[i]*fac[n-m-i]) % mod) % mod;
}
cout<<ans;
return ;
}