Problem:
Score : 600 points

Problem Statement

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).

Sample Input 1

1 1 2 2

Sample Output 1

14

For example, there are two paths from the point (0, 0) to the point (1, 1): (0,0) → (0,1) → (1,1) and (0,0) → (1,0) → (1,1), so f(1,1)=2.

Similarly, f(1,2)=3, f(2,1)=3, and f(2,2)=6. Thus, the sum is 14.

Sample Input 2

314 159 2653 589

Sample Output 2

602215194

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int MOD = 1e9+7;
const int maxm = 2e6+5;

LL F[maxm], Finv[maxm], inv[maxm];

LL comb(int n, int m) { //C(n, m)
    if(m < 0 || m > n) return 0;
    return F[n]*Finv[n-m]%MOD*Finv[m]%MOD;
}

void run_case() {
    int r1, c1, r2, c2, n;
    cin >> r1 >> c1 >> r2 >> c2;
    n = c2+r2+2;
    inv[1] = 1;
    for(int i = 2; i <= n; ++i)
        inv[i] = (MOD - MOD / i) * 1LL * inv[MOD % i] % MOD;
    F[0] = Finv[0] = 1;
    for(int i = 1; i <= n; ++i) {
        F[i] = F[i-1] *1LL* i % MOD;
        Finv[i] = Finv[i-1] * 1LL * inv[i] % MOD;
    }
    LL ans1=0,ans2=0,ans3=0,ans4=0;
    for(int i = 0; i <= r2; ++i) (ans1+=comb(c2+1+i,c2))%=MOD;
    for(int i = 0; i <= r2; ++i) (ans2+=comb(c1+i,c1-1))%=MOD;
    for(int i = 0; i <= r1-1; ++i) (ans3+=comb(c2+1+i,c2))%=MOD;
    for(int i = 0; i <= r1-1; ++i) (ans4+=comb(c1+i,c1-1))%=MOD;
    cout << (ans1+ans4-ans2-ans3+MOD)%MOD;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    //cout.setf(ios_base::showpoint);cout.precision(8);
    run_case();
    cout.flush();
    return 0;
}