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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const ll N=1e6+5,inf=1e18,mod=1e9+7;
#define test ll testcases;cin>>testcases; while(testcases--)
#define lp(var,start,end) for (ll var = start; var <end ; ++var)
#define rlp(var,start,end) for(ll var = start; var>=end ; var--)
#define pb push_back
#define mp make_pair
#define pf push_front
#define ff first
#define ss second
#define vll vector<ll>
#define vld vector<ld>
#define pll pair<ll,ll> 
#define pld pair<ld,ld> 
#define vpll vector<pll>
#define vpld vector<pld>
#define all(X) X.begin(),X.end()
#define sz(x) ((ll)((x).size()))
const ld pi=3.141592653589793238;

bool check_prime(ll n)
{
	ll flag=0,i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			flag=1;
			break;
		}
	}
	if(n==1)
	{
		return false;
	}
	else if(flag==0 || n==2 || n==3)
	{
		return true;
	}
	else
	{
		return false;
	}
}

ll fastexpo(ll a,ll b,ll MOD) 
{
	ll result=1;
	while(b>0)
	{
		if(b%2==1)
		{
			result=(result*a)%MOD;
		}
		a=(a*a)%MOD;
		b=b/2;
	}
	return result;
}

ll a[105];
ll dp[105][100005];

ll solve(ll n,ll k)
{
	if(k<0)
	{
		return 0;
	}
	if(dp[n][k]!=-1)
	{
		return dp[n][k];
	}
	ll ans;
	if(n>0 && k>0)
	{
		ans=(solve(n-1,k)%mod+solve(n,k-1)%mod-solve(n-1,k-1-a[n])%mod)%mod;
	}
	else
	{
		if(k<=a[n])
		{
			ans=1;
		}
		else
		{
			ans=0;
		}
	}
	dp[n][k]=ans;
	return ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll N,K,i,j;
    cin>>N>>K;
    for(i=0;i<N;i++)
    {
    	cin>>a[i];
    }
    for(i=0;i<105;i++)
    {
    	for(j=0;j<100005;j++)
    	{
    		dp[i][j]=-1;
    	}
    }
    cout<<solve(N-1,K)<<"\n";
	return 0;
}