Inclusion and exclusion of the original understanding of the general indeterminate equation --cf451E Classic title

/ * 
Given n boxes, the i-th box has ai flowers, now select from m flowers, Q number select programs 
by receiving repellent Theorem in m = X1 + X2 + .. + Xn 
C (m + n-. 1, N1) + SUM {(-1) ^ (m + n--l- (N1 + P *. 1 C) - (+ NP. 1), N1)}
    
 * / 
#include <bits / STDC ++ H.>
 the using  namespace STD;
 #define LL Long Long 
 #define MOD 1000000007 
LL ANS, n-, S, A [ 30 ], INV [ 30 ]; 

LL Pow (A LL, LL B) { 
    LL RES = . 1 ;
     the while (B) {
         IF (% B 2 ) 
            RES = RES * A% MOD; 
        B >> = . 1 ; A = A * A% MOD; 
    }
    return res;
}
ll C(ll y,ll x){
    if(y<0||x<0||y<x)return 0;
    y%=mod;
    if(y==0|| x==0)return 1;
    ll ans=1;
    for(ll i=0;i<x;i++)
        ans=(ll)ans*(y-i)%mod;
    for(ll i=1;i<=x;i++)
        ans=ans*inv[i]%mod;
    return ans;
}

int main(){
    cin>>n>>s;
    for(ll i=1;i<=20;i++)
        inv[i]=Pow(i,mod-2);
    
    for(int i=0;i<n;i++)cin>>A[i];
    
    for(ll i=0;i<(1<<n);i++){
        ll a=n-1,b=s+n-1,tmp=0;
        if(i==0){
            ans=(ans+C(b,a))%mod;
            continue;
        }
        for(int j=0;j<n;j++)
            if(i & ((ll)1<<j)){
                tmp++;
                b-=A[j]+1;
            }
        if(tmp%2){//减法 
            ans=(ans-C(b,a))%mod;
            ans=(ans+mod)%mod;
        }
        else {
            ans=(ans+C(b,a))%mod;
        } 
    }
    cout<<ans<<'\n';
}