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>
using namespace std;
using ll = long long;
const int maxn = 2e6 + 5;
const int P = 1e9 + 7;
vector<ll> fac;
vector<ll> inv;
vector<ll> facinv;
ll C(int n, int m) {
	++n;
	++m;
	return fac[n + m] * facinv[n] % P * facinv[m] % P;
}
int main() {
	fac.resize(maxn);
	facinv.resize(maxn);
	inv.resize(maxn);
	fac[0] = 1;
	inv[1] = 1;
	facinv[0] = 1;
	for(int i = 1; i < maxn; ++i) {
		fac[i] = fac[i - 1] * i % P;
		if(i != 1) {
			inv[i] = (P - P / i) * inv[P % i] % P;
		}
		facinv[i] = facinv[i - 1] * inv[i] % P;
	}
	int r1, c1, r2, c2;
	cin >> r1 >> c1 >> r2 >> c2;
	cout << ((C(r2, c2) - C(r1 - 1, c2) - C(r2, c1 - 1) + C(r1 - 1, c1 - 1)) % P + P) % P << '\n';
	return 0;
}