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 pb push_back
#define eb emplace_back
#define fi first
#define se second
#define endl '\n'
#define fcin ios::sync_with_stdio(false); cin.tie(nullptr);
#define inf 0x3f3f3f3f
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef pair<int, int> pii;

/* 
 * * stuff you should look for
 * * int overflow, array bounds
 * * special cases (n=1?), set tle
 * */

const int MOD = 1e9 + 7;
const int N = 3e6 + 6;
ll fat[N];

void init(){
    fat[0] = 1LL;
    for(ll i = 1; i < N; i++){
        fat[i] = fat[i-1] % MOD * (ll) i % MOD;
        fat[i] %= MOD;
    }
}

ll fast(ll a, ll p){
    if(p == 1) return a % MOD;
    if(p&1) return ((a % MOD) * (fast(a, p-1) % MOD)) % MOD;
    int x = p / 2;
    ll t = fast(a, x);
    return ((t % MOD) * (t % MOD)) % MOD;
}

ll nck(ll n, ll k){
    if(n < k) return 0;
    if(k == 0) return 1;
    ll up = fat[n];
    ll a = fast(fat[k], MOD - 2);
    ll b = fast(fat[n-k], MOD - 2);

    ll prod = up % MOD * a % MOD;
    prod %= MOD;
    prod *= b;

    return prod % MOD;
}

int main(){
    fcin;

    init();

    ll i, j, i2, j2;
    cin >> i >> j >> i2 >> j2;

    ll a = nck(i2 + j2 + 2, i2 + 1) - nck(i2 + j + 1, i2 + 1);
    ll b = nck(i + j2 + 1, i) - nck(i + j, i);

    cout << (a - b + MOD) % MOD << endl;
    return 0;
}