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>
typedef long long ll;

const int N=2000005,P=1000000007;
int a,b,c,d,Fac[N],Inv[N];

ll Pow(ll a,int b,ll s=1){for(;b;b>>=1,a=a*a%P)if(b&1)s=s*a%P;return s;}
inline ll Calc(int x,int y){return (ll)Fac[x+y]*Inv[x]%P*Inv[y]%P;}

int main()
{
    scanf("%d%d%d%d",&a,&b,&c,&d);
    int l=c+d+1;
    for(int i=Fac[0]=1;i<=l;++i)Fac[i]=(ll)Fac[i-1]*i%P;
    Inv[l]=Pow(Fac[l],P-2);
    for(int i=l;i>=1;--i)Inv[i-1]=(ll)Inv[i]*i%P;
    ll Ans=0;
    for(int i=a;i<=c;++i)Ans+=Calc(i+1,d)-Calc(i+1,b-1);
    return printf("%lld\n",(Ans%P+P)%P),0;
}