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 rep(i,s,n) for (int i = s; i < n; i++)
using namespace std;
using ll = long long;

const int MAX = 2100000;

ll fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit(ll p) {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  rep(i, 2, MAX) {
    fac[i] = fac[i - 1] * i % p;
    inv[i] = p - inv[p % i] * (p / i) % p;
    finv[i] = finv[i - 1] * inv[i] % p;
  }
}

// 二項係数計算
ll cmb(ll n, ll k, ll p) {
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % p) % p;
}

int main() {
  ll p = 1000000007;
  COMinit(p);
  ll r1,c1,r2,c2;
  cin >> r1 >> c1 >> r2 >> c2;
  ll a = 0;
  a = (cmb(r2 + c2 + 2, r2 + 1, p) + cmb(r1 + c1, r1, p) - cmb(r1 + c2 + 1, r1, p) - cmb(c1 + r2 + 1, c1, p)) % p;
  if (a < 0) a += p;
  cout << a << endl;
  return 0;
}
