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 <functional>
#include <iostream>
#include <iomanip>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <queue>
#include <numeric>
#include <bitset>
#include <cmath>

using namespace std;
using ll = long long;
using ull = unsigned long long;
// const ll mod =  998244353 ;
const ll mod =  1000000007 ;
const ll inf = 1e18;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1};


#define REP(i, n) for (ll i = 0; i < (n); ++i)
#define REP_FROM(i, j, n) for (ll i = (j); i < (n); ++i)
#define REP_REV(i, n) for (ll i = n-1; i >= 0; --i)
#define all(x) (x).begin(),(x).end()
#define sz(x) int(x.size())

template<typename T> void print_vec(vector<T> a) {int n = a.size();REP(i, n) {cout >> a[i]; if(i!=n-1) cout << " "; else cout << endl;}}
ll power(ll base, ll exponent){ if(exponent % 2){ return power(base, exponent - 1) * base % mod; }else if(exponent){ ll root_ans = power(base, exponent / 2); return root_ans * root_ans % mod; }else{ return 1; }}
ll inverse(ll x){ return power(x, mod - 2);}
ll gcd(ll a, ll b) {if(a < b) gcd(b, a);ll r;while(r=a%b) {a=b;b=r;}return b;}
template<typename T> ll sum(T begin, T end) {return accumulate(begin, end, 0ll);}
struct combination{ vector<ll> fact, inv; combination(int sz) : fact(sz + 1), inv(sz + 1){ fact[0] = 1; for(int i = 1; i <= sz; i++){ fact[i] = fact[i - 1] * i % mod; } inv[sz] = power(fact[sz], mod - 2); for(int i = sz - 1; i >= 0; i--){ inv[i] = inv[i + 1] * (i + 1) % mod; } } ll C(int p, int q) const{ if(q < 0 || p < q) return 0; return (fact[p] * inv[q] % mod * inv[p - q] % mod); } };

using Pair = pair<ll, ll>;

ll g(ll r, ll c, const combination& fact) {
    ll ans = 0;
    REP(i, r+1) {
        ans = (ans + fact.fact[i+c+1] * fact.inv[i+1] % mod * fact.inv[c] % mod) % mod;
    }
    return ans;
}

signed main() {
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll r1, c1, r2, c2;
    cin >> r1 >> c1 >> r2 >> c2;
    combination fact(r2+c2+10);
    ll ans = g(r2, c2, fact);
    ans = (ans - g(r1-1, c2, fact) + mod) % mod;
    ans = (ans - g(r2, c1-1, fact) + mod) % mod;
    ans = (ans + g(r1-1, c1-1, fact) + mod) % mod;
    cout << ans << endl;
    return 0;
}