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"

#define rep(i, a) for (int i = 0; i < (int)a; ++i)
#define REP(i, a, b) for (int i = (a); i < (int)b; ++i)

// debug methods
// usage: debug(x,y);
#define CHOOSE(a) CHOOSE2 a
#define CHOOSE2(a0, a1, a2, a3, a4, x, ...) x
#define debug_1(x1) cout << #x1 << ": " << x1 << endl
#define debug_2(x1, x2) \
  cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << endl
#define debug_3(x1, x2, x3)                                               \
  cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
       << x3 << endl
#define debug_4(x1, x2, x3, x4)                                           \
  cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
       << x3 << ", " #x4 << ": " << x4 << endl
#define debug_5(x1, x2, x3, x4, x5)                                       \
  cout << #x1 << ": " << x1 << ", " #x2 << ": " << x2 << ", " #x3 << ": " \
       << x3 << ", " #x4 << ": " << x4 << ", " #x5 << ": " << x5 << endl
#ifdef _DEBUG
#define debug(...)                                                      \
  CHOOSE((__VA_ARGS__, debug_5, debug_4, debug_3, debug_2, debug_1, ~)) \
  (__VA_ARGS__)
#else
#define debug(...)
#endif

using namespace std;
using ll = long long;
using P = pair<int, int>;

// auto mod int
const int mod = 1000000007;
struct mint {
  ll x;  // typedef long long ll;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};
// combination mod prime
struct combination {
  vector<mint> fact, ifact;
  combination(int n) : fact(n + 1), ifact(n + 1) {
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
  }
  mint operator()(int r, int c) {
    if (min(r, c) < 0) return 0;
    return fact[r + c] * ifact[r] * ifact[c];
  }
};

combination f(2e6);

void Main() {
  int r1, c1, r2, c2;
  cin >> r1 >> c1 >> r2 >> c2;
  auto g = [&](int r, int c) {
    mint x(0);
    return f(r + 1, c + 1) - f(0, c + 1);
  };
  mint ans = 0;
  ans += g(r2, c2);
  ans -= g(r2, c1 - 1);
  ans -= g(r1 - 1, c2);
  ans += g(r1 - 1, c1 - 1);
  cout << ans.x << endl;
}

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  Main();
  return 0;
}