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 <bits/stdc++.h>
#define MAXN 2000005
using namespace std;

const long long M=1e9+7;

int a,b,c,d;
long long inv[MAXN],fac[MAXN],invfac[MAXN],ans;

inline void fadd(long long &a,long long b){a+=b;if (a>=M) a-=M;return ;}

inline long long C(int n,int m)
{
	return fac[n]*invfac[m]%M*invfac[n-m]%M;
}

int main()
{
	inv[0]=inv[1]=invfac[0]=fac[0]=1LL;
	for (int i=1;i<MAXN;i++) fac[i]=fac[i-1]*i%M;
	for (int i=2;i<MAXN;i++) inv[i]=(M-M/i)*inv[M%i]%M;
	for (int i=1;i<MAXN;i++) invfac[i]=invfac[i-1]*inv[i]%M;
	scanf("%d %d %d %d",&a,&b,&c,&d);
	for (int i=b;i<=d;i++)
	{
		long long t=C(i+c+1,i+1);
		fadd(t,M-C(i+a,i+1));
		fadd(ans,t);
	}
	printf("%lld\n",ans);
	return 0;
}