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;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define ALL(A) A.begin(),A.end()
#define RALL(A) A.rbegin(),A.rend()
typedef long long LL;
typedef pair<LL,LL> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const LL mod=1000000007;
const LL LINF=1LL<<60;
const int INF=1<<30;
int dx[]={1,0,1,-1,1,-1,-1,0};
int dy[]={0,1,1,-1,-1,1,0,-1};

vector<LL> fact;
vector<LL> inver(2000001);
 
LL combi(int n,int r){
    if(n<r||n<0||r<0) return 0;
    return fact[n]%mod*inver[n-r]%mod*inver[r]%mod;
}
 
 
LL fpow(LL a, LL n){
    LL x = 1;
    while(n > 0){
        if(n&1){
            x=x*a%mod;
        }
        a=a*a%mod;
        n >>= 1;
    }
    return x;
}
 
void set_combi(){
    LL s=1;
    fact.push_back(1);
    for(int i=1;i<=2000000;i++){
        s*=i;
        s%=mod;
        fact.push_back(s);
    }
    inver[2000000]=fpow(fact[2000000],mod-2);
    for(int i=1999999;i>=0;i--){
        inver[i]=inver[i+1]*(i+1)%mod;
    }
}

LL f(LL r, LL c){
    return ((r + 2) * combi(r + c + 2, r + 2) %mod - c - 1 + mod) %mod * fpow(c + 1, mod - 2) %mod;
}

int main(){
    set_combi();
    LL r1,c1,r2,c2;cin >> r1 >> c1 >> r2 >> c2;
    assert((f(r2,c2) - f(r2,c1-1) %mod - f(r1-1,c2) %mod + f(r1-1,c1-1) + mod)%mod > 0);
    cout << (f(r2,c2) - f(r2,c1-1) %mod - f(r1-1,c2) %mod + f(r1-1,c1-1) + mod) %mod << endl;
    return 0;
}