#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007;

const ll MAX=2000010;
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;
}

int main() {
  ll r1,r2,c1,c2;
  cin>>r1>>c1>>r2>>c2;
  ll ans=0;
  COMinit();
  ans+=COM(r1+c1,r1);
  for(int i=r1+1;i<=r2;i++){
      ans+=COM(i+c2,i);
      ans%=MOD;
  }
  for(int i=c1+1;i<=c2;i++){
      ans+=COM(i+r2,i);
      ans%=MOD;
  }
  cout<<ans<<endl;
}
### Prompt

Please formulate a cpp solution to the following problem:

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).

Examples

Input

1 1 2 2


Output

14


Input

314 159 2653 589


Output

602215194

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007;

const ll MAX=2000010;
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;
}

int main() {
  ll r1,r2,c1,c2;
  cin>>r1>>c1>>r2>>c2;
  ll ans=0;
  COMinit();
  ans+=COM(r1+c1,r1);
  for(int i=r1+1;i<=r2;i++){
      ans+=COM(i+c2,i);
      ans%=MOD;
  }
  for(int i=c1+1;i<=c2;i++){
      ans+=COM(i+r2,i);
      ans%=MOD;
  }
  cout<<ans<<endl;
}
```