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
#include <bits/stdc++.h>
#include<iomanip>
#define ll long long
#define pb push_back
#define mp make_pair
#define MOD 1000000007
using namespace std;
 
ll f[2000002];
 
ll p(ll x, ll n) {
	ll res = 1;
	while(n) {
		if(n&1) {
			res=(res*x)%MOD;
		}
		x=(x*x)%MOD;
		n/=2;
	}
	return res;
}
 
ll nCr(ll n, ll r) {
	ll ans=1;
	ans=(ans*f[n])%MOD;
	ans=(ans*p(f[r],MOD-2))%MOD;
	ans=(ans*p(f[n-r],MOD-2))%MOD;
	return ans;
}
 
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll r1,c1,r2,c2,sum=0,i;
    f[0]=1; 
    for(i=1;i<2000002;i++) f[i]=(f[i-1]*i)%MOD;
    
	cin >> r1 >> c1 >> r2 >> c2;
	for(i=c1;i<=c2;i++) {
		ll temp=nCr(r2+1+i,r2)-nCr(r1+i,r1-1);
		temp = (temp+MOD)%MOD;
		sum=(sum+temp)%MOD;
	}
    cout << sum << "\n";
    return 0;
}