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: Accepted)

/*                        _
                       _oo0oo_
                      o8888888o
                      88" . "88
                      (| -_- |)
                      0\  =  /0
                   ____/`---'\____
                  /  \\|     |//  \
                 / \\||| /:\ |||// \
                / _||||| -:- |||||- \
               |   | \\\ \-/ /// |   |
               | \_|  ''\---/''  |_/ |
               \  .-\__  '-'  ___/-. /
            ____'. .'  /--.--\  `. .'_____
          /"" '<  `.___\_<|>_/___.' >' "" \
         | | :  `- \`.;`\ _ /`;.`/ - ` : | |
         \  \ `_.   \_ __\ /__ _/   .-` / _/
          `-.____`.___ \_____/___.-`___.-'

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 JENISH MONPARA S.    IIT PATNA    I_LOVE_ADITI_GOEL
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~*/
#include <bits/stdc++.h>
using namespace std;
const long double PI = 3.141592653589793;
const long double DEL = 1e-12;
const int mod =  1000000007;
const int LIM = 100005;


#define fio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define mpq priority_queue<int,vector<int>,greater<int>>
#define deb(a) cerr << #a << ": " << a << endl
#define ftt cin>>t;for(int tt=1;tt<=t;++tt)
#define forr(i,n) for((i)=1;(i)<=(n);(i)++)
#define Rev(v) reverse(v.begin(),v.end())
#define Sort(v) sort(v.begin(),v.end())
#define mem(a,b) memset(a,b,sizeof(a))
#define umii unordered_map<int,int>
#define all(a) a.begin() , a.end()
#define vvi vector<vector<int>>
#define pq priority_queue<int>
#define sqr(a) (((a) * (a)))
#define double long double
#define dbg cout<<"\nhi\n"
#define pii pair<int,int>
#define mii map<int,int>
#define vb vector<bool>
#define eb emplace_back
#define vi vector<int>
#define nl cout<<"\n"
#define pb push_back
#define int long long
#define sp <<" "<<
#define se second
#define fi first
 
int fpow(int x, int n){
    int res = 1;
    while(n){
          if(n&1){
                res = res * x % mod;
          }
          x = x * x % mod;
          n>>=1;
    }
    return res;
}
int gcd(int a,int b){
    if(b == 0)return a;
    return gcd(b,a % b);
}
void sieve(int n){
     bool prime[5*LIM];
     memset(prime, true, sizeof(prime));
      int rootn = (int)sqrt(n);
     for (int p = 2; p <= rootn; p++)
          if (prime[p] == true)
               for (int i = p*p; i <= n; i += p)
                    prime[i] = false;
     prime[1] = 0;

}

int cnt, sum, mid, mx = -1e17, mn = 1e17, a[2*LIM];
int n, m, d, t, i, j, k, l, r, x, y, z;
bool f, f1, f2;
string s;

//******************************************* CHECK CONSTRAINTS **************************************************


int32_t main()
{     fio;
      cin>>n>>k;
      forr(i,n){
            cin>>a[i];
      }
      vi dp(k+k+2,0),pre(k+k+2,0);
      for(i=0;i<=min(a[1],k);i++){
            dp[i]=1;
      }
      pre[0]=dp[0];
      for(i=1;i<=k;i++){
            pre[i]=pre[i-1]+dp[i];
            //cout<<pre[i]<<" ";
      }
      sum=a[1];
      for(i=2;i<=n;i++){
            sum+=a[i];
            dp.clear();
            for(j=0;j<=min(sum,k);j++){
                  if(j-a[i]-1<0){
                        dp[j]=pre[j]%mod;
                  }else
                        dp[j]=(pre[j]-pre[j-a[i]-1])%mod;
                  //dp[j]=max(dp[j],(int)0);
                  //cout<<dp[j]<<" ";
            }
            //nl;
            pre.clear();
            pre[0]=dp[0];
            for(l=1;l<=k;l++){
                  pre[l]=pre[l-1]+dp[l];
            }
      }
      cout<<dp[k];
     
}