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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<double,double> pdd;
#define SQ(i) ((i)*(i))
#define MEM(a, b) memset(a, (b), sizeof(a))
#define SZ(i) int(i.size())
#define FOR(i, j, k, in) for (int i=j ; i<(k) ; i+=in)
#define RFOR(i, j, k, in) for (int i=j ; i>=(k) ; i-=in)
#define REP(i, j) FOR(i, 0, j, 1)
#define REP1(i,j) FOR(i, 1, j+1, 1)
#define RREP(i, j) RFOR(i, j, 0, 1)
#define ALL(_a) _a.begin(),_a.end()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define X first
#define Y second
#ifdef jayinnn
#define TIME(i) Timer i(#i)
#define debug(...) do{\
    fprintf(stderr,"%s - %d (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<", ";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,deque<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
class Timer {
private:
    string scope_name;
    chrono::high_resolution_clock::time_point start_time;
public:
    Timer (string name) : scope_name(name) {
        start_time = chrono::high_resolution_clock::now();
    }
    ~Timer () {
        auto stop_time = chrono::high_resolution_clock::now();
        auto length = chrono::duration_cast<chrono::microseconds>(stop_time - start_time).count();
        double mlength = double(length) * 0.001;
        debug(scope_name, mlength);
    }
};
#else
#define TIME(i)
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0)
#endif

const ll MOD = 1000000007;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int iNF = 0x3f3f3f3f;
const ll MAXN = 100005;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

bool cmp(string a, string b){
    if(a.size() < b.size()) {
        return false;
    } else if(a.size() > b.size()){
        return true;
    } else {
        int size = a.size();
        for(int i=0;i<size;i++){
            if(a[i] < b[i]){
                return false;
            } else if(a[i] > b[i]){
                return true;
            }
        }
    }
    return false;
}
ll power(ll x, ll y){
    ll res = 1;
    x = x % MOD;

    while (y > 0) {
        if (y & 1){
            res = (res * x) % MOD;
        }

        y = y >> 1;
        x = (x * x) % MOD;
    }
    return res;
    
}
vector<ll> f, f_;
ll sol(ll x, ll y){
    ll ans = f[x + y];
    ans *= f_[x];
    ans %= MOD;
    ans *= f_[y];
    ans %= MOD;
    return ans;
}
/********** Good Luck :) **********/
int main () {
    TIME(main);
    IOS();
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    f.resize(c+d+5);
    f_.resize(c+d+5);
    f[1] = 1;
    for(int i=2;i<c+d+5;i++){
        f[i] = (f[i-1] * i) % MOD;
    }
    for(int i=1;i<c+d+5;i++){
        f_[i] = power(f[i], MOD - 2);
    }
    ll ans = 0;
    // debug(f);
    ans += sol(c + 1, d + 1) - 1;
    ans -= sol(c + 1, b) - 1;
    ans -= sol(a, d + 1) - 1;
    ans += sol(a, b) - 1;
    while (ans < 0) ans += MOD;
    ans %= MOD;
    cout << ans << endl;
    return 0;
}