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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; i++)
#define repo(i, n) for(int i=1; i<=n; i++)
#define ssort(a) sort(a.begin(), a.end())
#define INF 1001001001
#define INFll 100100100100100
// debug用
#define PrintVec(x) for (auto elementPrintVec: x) { cout << elementPrintVec << " "; } cout << endl;

using namespace std;
using ull = unsigned long long;
using ll = long long;
using P = pair<int, int>;
const int mod = 1000000007;


#include <iostream>
using namespace std;
/*
1≤k≤n≤10^7
p は素数 (上の実装ではさらに p>n を仮定している)
*/

const int MAX = 2000100;//階乗の配列数
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(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
  if(a<n)return 0;
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}


int main(){
  int r1, c1, r2, c2;
  cin >> r1 >> c1 >> r2 >> c2;
  COMinit();
  ll ans=0;
  for(int x=r1; x<=r2; x++){
    ans+=COM((ll)(x+1+c2), (ll)c2)-COM((ll)(x+c1), (ll)(x+1));
    ans+=mod;
    ans%=mod;
  }
    cout << ans << endl;
  return 0;
}