/*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#include #include #include #include #include #include #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 = 100050;const ll mod = 1000000007LL;ll read(){ ll x=0,f=1; char ch=getchar(); while(!(ch>='0'&&ch<='9')){ if(ch=='-')f=-1;ch=getchar();}; while(ch>='0'&&ch<='9'){x=x*10+(ch-'0');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 = 1; while(b){ if(b&1) ret = (ret*a) % mod; a = (a*a) % mod; b >>= 1; } return ret;}inline ll inv(ll x){ return q_pow(x,mod-2);}int main(){ freopen("problem.in","r",stdin); freopen("problem.out","w",stdout); n = read();m = read(); fo(i,1,m) x[i] =read(),y[i]=read(); fo(i,1,m){ usd[x[i]] = true; usd[y[i]] = true; } int t = n; fo(i,1,n) if(usd[i]) t--; fac[0] = fac[1] = 1; fo(i,2,n) fac[i] = (fac[i-1]*i) % mod; c[0] = c[t] = 1; if(t > 0) c[1] = c[t-1] = t; fo(i,2,t-2)c[i] = ((c[i-1]*(t-i+1)%mod)*inv(i))%mod; ans = fac[n-m]; fo(i,1,t){ if(i&1) ans = (ans + 100*mod - (c[i]*fac[n-m-i]) % mod) % mod; else ans = (ans + (c[i]*fac[n-m-i]) % mod) % mod; } cout<