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>
#include <cstring>
#define int long long
int inv[2000011],fac[2000011],finv[2000011],r1,r2,c1,c2;const int p=1e9+7;
int qpow(int bs,int ex){int ans=1;while(ex){if(ex&1)ans=ans*bs%p;bs=bs*bs%p;ex>>=1;}return ans;}
int c(int n,int m){if(n>m)return 0;return fac[m]*finv[n]%p*finv[m-n]%p;}
int sum(int n,int k){return (n-k+1)*qpow(k+1,p-2)%p*c(k,n+1)%p;}
signed main()
{
	scanf("%lld%lld%lld%lld",&r1,&c1,&r2,&c2);
	inv[1]=1;for(int i=2;i<=2000010;++i)inv[i]=(p*inv[p%i]-p/i*inv[p%i])%p;fac[0]=1;
	for(int i=1;i<=2000010;++i)fac[i]=fac[i-1]*i%p;finv[0]=1;for(int i=1;i<=2000010;++i)finv[i]=finv[i-1]*inv[i]%p;int ans=0;
	for(int i=r1;i<=r2;++i)ans+=((sum(i+c2,i)-sum(i+c1-1,i))%p+p)%p,ans%=p;printf("%lld\n",ans);return 0;
}