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 <iostream>
using namespace std;

const int MAX=3000005;
const int MOD=1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

void f() {
    fac[0]=1;
    fac[1]=1;
    finv[0]=1;
    finv[1]=1;
    inv[1]=1;
    for (int i=2; i<MAX; i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}

long long C(int n, int k){
    if (n<k) return 0;
    if (n<0 || k<0) return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

long long F(int x, int y) {
    return (C(x+y+2, x+1)-x-y-2)%MOD;
}

int main(void) {
    int r1, c1, r2, c2;
    long long X;
    scanf("%d%d%d%d", &r1, &c1, &r2, &c2);
    f();
    X=F(r2, c2)-F(r1-1, c2)-F(r2, c1-1)+F(r1-1, c1-1);
    // X=F(r2, c2)-F(0, c2)-F(r2, 0)+F(0, 0);
    if (X<0) X+=2*MOD;
    X%=MOD;
    printf("%lld\n", X);
    return 0;
}
