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;
using ll = long long;
#define REP(i, a, b) for(int i = a, i##_len = (b); i < i##_len; ++i)

const int MAX = 2000100;
const int MOD = 1000000007;

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

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = 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;
    }
}

// 二項係数計算
long long COM(long long n, long long k){
    if (n < k) return 0LL;
    if (n < 0LL || k < 0LL) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long f(ll r, ll c) {
    return COM(c+r, c);
}

int main()
{
    ll r1, c1, r2, c2;
    scanf("%lld %lld %lld %lld", &r1, &c1, &r2, &c2);

    COMinit();

    //ll res = f(r2+1, c2+1) - f(r1,c2+1) - (f(r2+1, c1) - f(r1, c1));
    //printf("%lld\n", res%MOD);
    ll res = (f(r2+1, c2+1) - f(r1, c2+1) + MOD)%MOD;
    res = (res - (f(r2+1, c1) - f(r1,c1) + MOD)%MOD + MOD)%MOD;
    printf("%lld\n", res);
    return 0;
}