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: Accepted)

#include<bits/stdc++.h>
#define ll long long
#define re register
using namespace std;
const int mod=1e9+7;
const int maxn=100+10;
const int maxk=1e5+10;

template<typename T> void read(T &x){
	T f=1;x=0;
	char c=getchar();
	while(c<'0' || c>'9'){if(c=='-') f=-1;c=getchar();}
	while(c>='0' && c<='9'){x=(x<<3)+(x<<1)+c-48;c=getchar();}
	x*=f;
}
ll n,k;
ll a[maxn];
ll f[maxn][maxk];
ll pre_sum[maxn];
ll pre_ans[maxn][maxk];
int main(){
	read(n);read(k);
	for(re int i=1;i<=n;i++){
		read(a[i]);
		pre_sum[i]+=pre_sum[i-1]+a[i];
	} 
	f[0][0]=1;
	for(re int i=0;i<=k;i++){
		pre_ans[0][i]=1;
	}
	for(re int i=0;i<=n;i++) f[i][0]=pre_ans[i][0]=1;
	/*
	for(re int i=1;i<=n;i++){
		for(re int j=0;j<=k && j<=pre_sum[i];j++){
			for(re int t=0;t<=a[i];t++){
				if(j-t>=0) f[i][j]=(f[i][j]%mod+f[i-1][j-t]%mod)%mod;
			}
		}
	}
	*/	
	for(re ll i=1;i<=n;i++){
		for(re ll j=0;j<=k;j++){
			if(a[i]>=j) f[i][j]=(pre_ans[i-1][j])%mod;
			else f[i][j]=(f[i][j]+pre_ans[i-1][j]-pre_ans[i-1][j-a[i]-1])%mod;
			if(j>0) pre_ans[i][j]+=f[i][j]+pre_ans[i][j-1];
			else pre_ans[i][j]=f[i][j];
		}
	}
	//cout<<pre_ans[n][k]<<endl;
	cout<<f[n][k]<<endl;
	return 0;
}