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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ff first
#define ss second
#define int ll
#define pb push_back
#define setbits(x)          __builtin_popcountll(x)
#define endl "\n"
typedef long long ll;
const int mod=1000000007;
const int mx=2000010;
int fact[2000010];
ll give(ll x, ll y, ll p)
{ll res = 1; x = x % p;
while (y > 0)
{if(y & 1) res = (res*x) % p;
y = y>>1;x = (x*x) % p;
}return res;
}
int find(int n,int m)
{
    if(n<0||m<0)
return 0;
    int ans=fact[n+m];
    ans*=give(fact[n],mod-2,mod);
    ans%=mod;
    ans*=give(fact[m],mod-2,mod);
    ans%=mod;
    return ans;
}
int32_t main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);

fact[0]=1;
for(int i=1;i<mx;i++)
{
   fact[i]=fact[i-1]*i;
   fact[i]%=mod;
}

int r1,r2,c1,c2;

cin>>r1>>c1>>r2>>c2;
r1-=1;
r2-=1;
c1-=1;
c2-=1;
int ans=0;
/// r2,c2;
for(int i=1;i<=r2+1;i++)
{
    ans+=find(i,c2);
    ans%=mod;
}
/// r1,c2

for(int i=1;i<=r1;i++)
{
    ans-=find(i,c2);
    ans+=mod;
    ans%=mod;
}
/// r2,c1
for(int i=1;i<=r2+1;i++)
{
    ans-=find(i,c1);
    ans+=mod;
    ans%=mod;
}

// r1-1
for(int i=1;i<=r1;i++)
{
    ans+=find(i,c1-1);
    ans+=mod;
}

cout<<ans<<endl;
    return 0;
}