Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.

Let us define a function f(r, c) as follows:

* f(r,c) :=  (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)



Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).

Constraints

* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


r_1 c_1 r_2 c_2


Output

Print the sum of f(i, j) modulo (10^9+7).

Examples

Input

1 1 2 2


Output

14


Input

314 159 2653 589


Output

602215194
#include <cstdio>

typedef long long ll;
const int maxn=2000000+10,mod=1000000007;
ll fac[maxn],ifac[maxn];

ll power(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1)ret=ret*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ret;
}

ll inv(ll a){
	return power(a,mod-2);
}

ll C(ll n,ll m){
	return fac[n]*ifac[m]%mod*ifac[n-m]%mod;
}

int main(){
	// freopen("in.txt","r",stdin);
	fac[0]=ifac[0]=1;
	for(int i=1;i<maxn;++i)fac[i]=fac[i-1]*i%mod;
	ifac[maxn-1]=inv(fac[maxn-1]);
	for(int i=maxn-2;i>=1;--i)ifac[i]=ifac[i+1]*(i+1)%mod;
	int r1,c1,r2,c2;
	scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
	printf("%lld\n",(C(r2+c2+2,r2+1)+C(r1+c1,r1)-C(r2+c1+1,r2+1)-C(r1+c2+1,c2+1)+mod+mod)%mod);
	// printf("%lld %lld %lld %lld\n",C(r2+c2+2,r2+1),C(r1+c1+2,r1+1),C(r2+c1+1,r2+1),C(r1+c2+1,c2+1));
	return 0;
}