/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=2e6,p=1e9+7;
int fac[N+10],inv[N+10];
int X[10],Y[10],Ans;
void prepare(){
	fac[0]=inv[0]=inv[1]=1;
	for (int i=1;i<=N;i++)	fac[i]=1ll*i*fac[i-1]%p;
	for (int i=2;i<=N;i++)	inv[i]=1ll*(p-p/i)*inv[p%i]%p;
	for (int i=1;i<=N;i++)	inv[i]=1ll*inv[i-1]*inv[i]%p;
}
struct S1{
	int x,y,k;
	void insert(int _x,int _y,int _k){x=_x,y=_y,k=_k;}
}A[4],B[4];
int F(int x1,int y1,int x2,int y2){return 1ll*fac[x2-x1+y2-y1]*inv[x2-x1]%p*inv[y2-y1]%p;}
int solve(const S1 &a,const S1 &b){
	int x1=a.x,y1=a.y,x2=b.x,y2=b.y,res=0;
	for (int i=X[3];i<=X[4];i++)	res=(res+1ll*F(x1,y1,i,Y[3]-1)*F(i,Y[3],x2,y2)%p*(-i-Y[3])%p)%p;
	for (int i=Y[3];i<=Y[4];i++)	res=(res+1ll*F(x1,y1,X[3]-1,i)*F(X[3],i,x2,y2)%p*(-i-X[3])%p)%p;
	for (int i=X[3];i<=X[4];i++)	res=(res+1ll*F(x1,y1,i,Y[4])*F(i,Y[4]+1,x2,y2)%p*(i+Y[4]+1)%p)%p;
	for (int i=Y[3];i<=Y[4];i++)	res=(res+1ll*F(x1,y1,X[4],i)*F(X[4]+1,i,x2,y2)%p*(i+X[4]+1)%p)%p;
	return res;
}
int main(){
	prepare();
	for (int i=1;i<=6;i++)	X[i]=read();
	for (int i=1;i<=6;i++)	Y[i]=read();
	A[0].insert(X[1]-1,Y[1]-1,1),A[1].insert(X[1]-1,Y[2],-1),A[2].insert(X[2],Y[1]-1,-1),A[3].insert(X[2],Y[2],1);
	B[0].insert(X[6]+1,Y[6]+1,1),B[1].insert(X[6]+1,Y[5],-1),B[2].insert(X[5],Y[6]+1,-1),B[3].insert(X[5],Y[5],1);
	for (int i=0;i<4;i++)	for (int j=0;j<4;j++)	Ans=(Ans+1ll*A[i].k*B[j].k*solve(A[i],B[j])%p)%p;
	printf("%d\n",(Ans+p)%p);
	return 0;
}