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>
//ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll mod3=1000003;
ll mod4=998244853;
ll inf=1LL<<60;
long double pi=3.14159265358979323846L;
double eps=1e-12;
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,n,m) for(ll i=n;i>=m;i--)
#define srep(itr,st) for(auto itr=st.begin();itr!=st.end();itr++)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll gcd(ll a,ll b){
    if(a<b)swap(a,b);
    if(b==0)return a;
    if(a%b==0)return b;
    return gcd(b,a%b);
}
ll Pow(ll n,ll k){
    ll ret=1;
    ll now=n;
    while(k>0){
        if(k&1)ret*=now;
        now*=now;
        k/=2;
    }
    return ret;
}
ll beki(ll n,ll k,ll md){
  ll ret=1;
  ll now=n;
  while(k>0){
    if(k%2==1){
      ret*=now;
      ret%=md;
    }
    now*=now;
    now%=md;
    k/=2;
  }
  return ret;
}
ll gyaku(ll n,ll md){
  return beki(n,md-2,md);
}
ll gya[2000110];
ll kai[2000110];
void nckinit(ll n,ll md){
  kai[0]=1;
  kai[1]=1;
  for(int i=2;i<=n;i++){
    kai[i]=kai[i-1]*i;
    kai[i]%=md;
  }
  gya[n]=gyaku(kai[n],md);
  for(int i=n-1;i>=1;i--){
    gya[i]=gya[i+1]*(i+1);
    gya[i]%=md;
  }
  gya[0]=1;
}
ll nck(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0||n==k)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  ret*=gya[k];
  ret%=md;
  return ret;
}
ll npk(ll n,ll k,ll md){
  if(k<0)return 0;
  if(k==0)return 1;
  ll ret=kai[n];
  ret*=gya[n-k];
  ret%=md;
  return ret;
}
ll s(ll r,ll c){
    ll ret=0;
    rep(i,1,c+2){
        (ret+=nck(r+i,r,mod))%=mod;
    }
    return ret;
}
int main(){
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    ll r1,c1,r2,c2;
    cin>>r1>>c1>>r2>>c2;
    nckinit(2000010,mod);
    ll ans=0;
    ans+=s(r2,c2);
    (ans+=-s(r1-1,c2)+mod)%=mod;
    (ans+=-s(r2,c1-1)+mod)%=mod;
    (ans+=s(r1-1,c1-1)+mod)%=mod;
    cout<<ans<<endl;
}