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: Accepted)

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <unordered_map>
#include <vector>
#include <string.h>
#include <set>

using namespace std;

#define COUT(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << endl

template<class T> void printvec(const vector<T>& v) {
  for (auto x : v) { cout << x << " "; }
  cout << endl;
}
template<class T> void printtree(const vector< vector<T> >& tree) {
  for (long long i = 0; i < tree.size(); ++i) {
    cout << i + 1 << ": "; printvec(tree[i]);
  }
}
template<class T, class U> void printmap(const map<T, U>& mp) {
  for (auto x : mp) { cout << x.first << "=>" << x.second << endl; }
}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

#define rep(i, n) for(ll i = 0; i < n; ++i)
#define all(s) s.begin(), s.end()
#define fr first
#define sc second
#define mp make_pair
#define pb push_back

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> triple;
typedef double D;

const ll INF = 1e9;
const ll MOD = 1000000007;  // 1e9 + 7

// Mod int
// cf. https://www.youtube.com/watch?v=1Z6ofKN03_Y
struct mint {
  ll x;
  mint(ll x = 0) : x((x + MOD) % MOD) {}
  mint& operator+= (const mint a) {
    if ((x += a.x) >= MOD) x %= MOD;
    return *this;
  }
  mint operator+ (const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint& operator-= (const mint a) {
    if ((x += MOD - a.x) >= MOD) x %= MOD;
    return *this;
  }
  mint operator- (const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint& operator*= (const mint a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  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.
// cf. https://www.youtube.com/watch?v=1Z6ofKN03_Y
struct Combination {
  vector<mint> fact, ifact;
  Combination(int n) { init(n); }
  void init(int n) {
    assert(n < MOD); // n must be lower than MOD.

    fact.resize(n + 1);
    ifact.resize(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 n, int k) const {
    if (k < 0 || k > n) { return 0; }
    return fact[n] * ifact[k] * ifact[n-k];
  }
  mint perm(int n, int k) const {
    if (k < 0 || k > n) { return 0; }
    return fact[n] * ifact[n-k];
  }
};

// int main(int argc, char** argv) {
//   MOD = 13;
//
//   Combination c(12);
//   cout << c(12, 0).x << endl;  // 1  = 1 % 13
//   cout << c(12, 1).x << endl;  // 12 = 12 % 13
//   cout << c(12, 2).x << endl;  // 1  = 66 % 13 = (12 * 11 / 2) % 13
//   cout << c(12, 3).x << endl;  // 12 = 220 % 13 = (12 * 11 * 10 / (3 * 2 * 1)) % 13
//   cout << c(12, 11).x << endl; // 12 = 12 % 13
//   cout << c(12, 12).x << endl; // 1  = 12 % 13
//
//   cout << c.perm(12, 3).x << endl; // 7 = 12 * 11 * 10 % 13
// }

// r+1+c+1 <= 2000005
Combination comb(2000005);

// cummulated sum of points in [0, r] and [0, c].
mint g(int r, int c) {
  // the result is "the value in (r+1,c+1)" - 1.
  return comb(r+c+2, r+1) - 1;
}

int main(int argc, char** argv) {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  //cout << setprecision(10) << fixed;

  ll r1, c1, r2, c2;
  cin >> r1 >> c1 >> r2 >> c2;

  mint ans = g(r2, c2);
  ans -= g(r1-1,c2);
  ans -= g(r2,c1-1);
  ans += g(r1-1,c1-1);
  cout << ans.x << endl;
}