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>

#define ll long long
#define fornum(A,B,C) for(A=B;A<C;++A)
#define pii pair<int,int>
#define pll pair<ll,ll>

using namespace std;

/////////////////////////////////////////////////////
#define MOD (ll)(1e9+7)
ll R1, C1, R2, C2;
ll i,j,k;
ll Fa[2020202];

ll moddiv(ll a,ll b){
    a %= MOD;
    b %= MOD;
    for(ll m=MOD-2;m>0;m/=2){
        if(m&1){
            a=a*b%MOD;
        }
        b=b*b%MOD;
    }
    return a;
}

ll f(const ll& a,const ll& b){
    return moddiv(Fa[a + b], Fa[a] * Fa[b]);
    
}
int main(){
    scanf("%lld%lld%lld%lld", &R1, &C1, &R2, &C2);
    Fa[0] = 1;
    fornum(i,0,2010101){
        Fa[i + 1] = Fa[i] * (i + 1) % MOD;
    }
    ll ans = 0;
    
    ans = f(R2 + 1, C2 + 1) - f(R1, C2 + 1) - f(R2 + 1, C1) + f(R1, C1);
    ans = (ans + MOD * MOD) % MOD;
    printf("%lld", ans);


    return 0;
}