Problem:
Score : 100 points

Problem Statement

There are N children, numbered 1, 2, \ldots, N.

They have decided to share K candies among themselves.
Here, for each i (1 \leq i \leq N), Child i must receive between 0 and a_i candies (inclusive).
Also, no candies should be left over.

Find the number of ways for them to share candies, modulo 10^9 + 7.
Here, two ways are said to be different when there exists a child who receives a different number of candies.

Constraints

All values in input are integers.

1 \leq N \leq 100

0 \leq K \leq 10^5

0 \leq a_i \leq K

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \ldots a_N

Output

Print the number of ways for the children to share candies, modulo 10^9 + 7.

Sample Input 1

3 4
1 2 3

Sample Output 1

5

There are five ways for the children to share candies, as follows:

(0, 1, 3)

(0, 2, 2)

(1, 0, 3)

(1, 1, 2)

(1, 2, 1)

Here, in each sequence, the i-th element represents the number of candies that Child i receives.

Sample Input 2

1 10
9

Sample Output 2

0

There may be no ways for the children to share candies.

Sample Input 3

2 0
0 0

Sample Output 3

1

There is one way for the children to share candies, as follows:

(0, 0)

Sample Input 4

4 100000
100000 100000 100000 100000

Sample Output 4

665683269

Be sure to print the answer modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define PI 3.141592653589793238462643383279502884L
#define make_unique(vec) vec.resize(distance(vec.begin(),unique(vec.begin(),vec.end())))
#define Sort(vec) sort(vec.begin(),vec.end())
#define RSort(vec) sort(vec.rbegin(),vec.rend())
#define endl "\n"
#define tr1(i) cout<<i<<endl;
#define tr2(i,j) cout<<i<<" "<<j<<endl;
#define tr3(i,j,k) cout<<i<<" "<<j<<" "<<k<<endl;
#define vvi vector<vector<int> > 
#define vvl vector<vector<ll> >
#define all(st) st.begin(),st.end()
#define vpl vector<pair<ll,ll> >
#define vpi vector<pair<int,int> >
#define pi pair<int,int>
#define pl pair<ll,ll>
#define al vector<list<int> >
#define vs vector<string>
#define vb vector<bool>
#define vi vector<int>
#define vl vector<ll>
#define vc vector<char>
#define rep(i,l,r) for(int i=l;i<r;i++)
#define repit(st) for(auto it=st.begin();it!=st.end();it++)
#define ull unsigned long long int
#define eb emplace_back
#define pb push_back
#define ll long long int
#define minf -(1e18)
#define inf 1e18
#define f first
#define se second
#define mod 1000000007
#define mx 100001
using namespace std;
ll mi(ll n,ll m){ll pw=n%m;ll ex=m-2;ll ans=1;while(ex){if(ex&1) ans = ans*pw%m;pw = pw*pw%m;ex>>=1;}return ans%mod;}
ll pw(ll a,ll n){ll pw=a%mod;ll ex=n;ll ans=1;while(ex){if(ex&1) ans = ans*pw%mod;pw = pw*pw%mod;ex>>=1;}return ans%mod;}
template<typename T>istream &operator>>(istream &is, vector<T> &v) { for (T &t : v) is >> t; return is; }
template<typename T>ostream &operator<<(ostream &os, const vector<T> &v) { for (const T &t : v) {os << t <<" ";} os << '\n'; return os; }
void show(ll x){cout<<x<<" ";}
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll max(ll a,ll b){return a>b?a:b;}
ll min(ll a,ll b){return a<b?a:b;}
ll Abs(ll x){return x>0?x:(-x);}
ll Ceil(ll a,ll b){return a/b+(a%b!=0);}
#define bs binary_search
#define lb lower_bound
#define ub upper_bound
#define mkp make_pair
//#define d double
void solve(){
	ll n,k;
	cin>>n>>k;
	vl a(n);
	cin>>a;
	vl dp(k+1);
	for(int i=0;i<=a[0];i++) dp[i]=1;
	for(int i=1;i<n;i++){
		vl tmp=dp;
		vl pre(k+1);
		pre[0]=dp[0];
		for(int j=1;j<=k;j++) pre[j]=(pre[j-1]+dp[j])%mod;
		for(int l=1;l<=k;l++){
			if(a[i]>=l) tmp[l]=(tmp[l]+pre[l-1])%mod;
			else tmp[l]=(tmp[l]+pre[l-1]-pre[l-a[i]-1]);
		}
		dp=tmp;
	}
	cout<<dp[k];
	
	
 
 
 
}
int32_t main(){
int t=1;
//cin>>t;
while(t--){
	solve();
}
}