#include <bits/stdc++.h>

using namespace std;
// #include <atcoder/all>
// using namespace atcoder;
using ll =long long;
typedef pair<int,int> P;
#define SORT(a) sort((a).begin(),(a).end())
#define REV(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define debug(x)  cerr << #x << " = " << (x) << endl;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

void coY() {cout <<"Yes"<<endl;}
void coN(){cout <<"No"<<endl;}

const int dy[] = {0,0,1,-1};
const int dx[] = {1,-1,0,0};
const ll mod = 1e9+7;
const ll MOD = 998244353;
const double PI=3.14159265358979323846;
const int INF = 1001001001;
//Write From this Line

// now += mint(2).pow(s) - 1; こんな感じで、2のs乗を求められrう
// auto mod int
// https://youtu.be/L8grWxBlIZ4?t=9858
// https://youtu.be/ERZuLAxZffQ?t=4807 : optimize
// https://youtu.be/8uowVvQ_-Mo?t=1329 : division
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint operator-() const { return mint(-x);}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }

  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
int main()
{
	int n, m;
	cin >>  n >> m;
	vector<int> a(n);
	rep(i,n) cin >>  a[i];
	ll sum = 0;
	rep(i,n){
		sum += a[i];
	}
	mint ans = 1;
	for(ll i = 0; i < sum+n; i++){
		ans *= (m+n-i);
		ans /= (i+1);
	}
	cout << ans << endl;


}

#include <iostream>
#include <string>
using namespace std;
int main(){
int m,h;cin>>m>>h;
string  ans  = "Yes";
if(h%m)ans = "No";
cout<<ans<<endl;

return 0;
}
