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++ (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define reg register
#define ll long long
using namespace std;
const int MAXN = 2e6 + 10;
const int mod = 1e9 + 7;
ll r1, c1, r2, c2;
ll fac[MAXN], inv[MAXN];
ll fastpow(ll a, ll b) {
    ll ret = 1;
    while(b) {
        if(b & 1)
            ret = ret * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ret;
}
ll getC(ll a, ll b) {
    return (a < b ? 0 : (fac[a] * inv[b] % mod * inv[a - b] % mod));
}
int main() {
    scanf("%lld%lld%lld%lld", &r1, &c1, &r2, &c2);
    ll n = max(c1, c2) + r2 + 2;
    fac[0] = inv[0] = 1;
    for(reg int i = 1; i <= n; ++i)
        fac[i] = fac[i - 1] * i % mod;
    inv[n] = fastpow(fac[n], mod - 2);
    for(reg int i = n; i; i--)
        inv[i - 1] = inv[i] * i % mod;
    ll ret1 = 0, ret2 = 0;
    for(reg int i = r1; i <= r2; ++i)
        ret1 = (ret1 + getC(i + c2 + 1, i + 1)) % mod;
    for(reg int i = r1; i <= r2; ++i)
        ret2 = (ret2 + getC(i + c1, i + 1)) % mod;
    printf("%lld\n", (ret1 - ret2 + mod) % mod);
    return 0;
}