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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

// 素数 p で割ったあまりに関する構造体 (特に p = 1000000007 とすることが多い)
template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) v += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr istream& operator >> (istream &is, Fp<MOD>& x) noexcept {
        return is >> x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};

// 二項係数ライブラリ
template<class T> struct BiCoef {
    vector<T> fact_, inv_, finv_;
    constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        int MOD = fact_[0].getmod();
        for(int i = 2; i < n; i++){
            fact_[i] = fact_[i-1] * i;
            inv_[i] = -inv_[MOD%i] * (MOD/i);
            finv_[i] = finv_[i-1] * inv_[i];
        }
    }
    constexpr T com(int n, int k) const noexcept {
        if (n < k || n < 0 || k < 0) return 0;
        return fact_[n] * finv_[k] * finv_[n-k];
    }
    constexpr T fact(int n) const noexcept {
        if (n < 0) return 0;
        return fact_[n];
    }
    constexpr T inv(int n) const noexcept {
        if (n < 0) return 0;
        return inv_[n];
    }
    constexpr T finv(int n) const noexcept {
        if (n < 0) return 0;
        return finv_[n];
    }
};

int main() {
    // 1000000007 で割ったあまりのクラス
    using mint = Fp<1000000007>;

    // 二項係数計算の前処理 (nCk, n, k < 101010 について O(1) でとれるようにする)
    BiCoef<mint> bc(2000010);

    // スターリング数
    int r1, c1, r2, c2;
    cin >> r1 >> c1 >> r2 >> c2;

    mint res = 0;
    for(int i=r1; i<=r2; ++i){
        for(int j=c1; j<=c2; ++j){
            mint tmp = bc.fact(i+j) *bc.finv(i) * bc.finv(j);
            res += tmp;
        }
    }

    cout << res << endl;

}