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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll MOD = 1000000007;

ll factorial[2000008];
ll iFactorial[2000008];

ll bigMode( ll base, ll b ){
    ll ans = 1;
    while( b ){
        if( b%2 == 1 ){
            ans = (ans*base)%MOD;
        }
        base = (base*base)%MOD;
        b /= 2;
    }
    return ans%MOD;
}

void fun(  ){
    factorial[0] = 1;
    iFactorial[0] = 1;
    for( int i = 1; i <= 1000000; i++ ){
        factorial[i] = (i*factorial[i-1])%MOD;
        iFactorial[i] = (iFactorial[i-1]*bigMode(i, MOD-2))%MOD;
    }
}

int main(){
    fun();
    int r1,r2,c1,c2;
    while( cin >> r1 >> c1 >> r2 >> c2 ){
        ll ans = 0;
        for( int i = r1 ; i <= r2; i++ ){
            int r = i+c2+1;
            int c = c2;
            if( r < c )swap( r,c );
            ll tempAns = factorial[r];
            tempAns = (tempAns*iFactorial[c])%MOD;
            tempAns = (tempAns*iFactorial[r-c])%MOD;
            ans = (ans+tempAns)%MOD;
            r = i+c1;
            c = c1-1;
            if( r < c )swap(r,c);
            tempAns = factorial[r];
            tempAns = (tempAns*iFactorial[c])%MOD;
            tempAns = (tempAns*iFactorial[r-c])%MOD;
            ans = ans - tempAns;
            if( ans < 0 ){
                ans = ans+MOD;
            }
            ans = ans%MOD;
        }
        cout << ans%MOD << endl;
    }
}
/*
7 2
0 1 0 0 0 1 0
*/