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)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author P.Alizadeh
 */

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long int   ll;
typedef long double     ld;
typedef pair<ll,ll>     pll;
typedef complex<ld>     cld;

#define all(x)          x.begin(),x.end()
#define Sort(x)         sort(all(x))
#define kill(x)         return cout << x << endl, 0;
#define Mp              make_pair
#define X               first
#define Y               second
#define sep             " "
#define endl            "\n"
#define fastio          ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io         freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define ld_io           cout << setprecision(10) << fixed;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll pw(ll a, ll b, ll md) {
    if (!b)
        return 1;
    if (b & 1)
        return a * pw(a * a % md, b / 2, md) % md;
    return pw(a * a % md, b / 2, md) % md;
}

const ll MAXN = 1e5 + 10,
         INF  = 8e18,
         MOD  = 1e9 + 7,
         LOG  = 32;
//const ld PI   = M_PI;

ll n, k, dp[110][MAXN], a[110];

int main()
{
    fastio;
    //ld_io;

    cin >> n >> k;
    for (ll i = 1; i <= n; i++)
        cin >> a[i];

    dp[0][0] = 1;
    for (ll i = 1; i <= n; i++) {
        for (ll j = 0; j < MAXN; j++) {
            if (j == 0)
                dp[i][j] = 1;
            else if (j <= a[i])
                dp[i][j] = (dp[i][j-1] + dp[i-1][j]) % MOD;
            else
                dp[i][j] = (dp[i][j-1] - dp[i-1][j-a[i]-1] + dp[i-1][j]) % MOD;
        }
    }

    if (dp[n][k] >= 0)
        cout << dp[n][k] % MOD << endl;
    else
        cout << (dp[n][k] % MOD) + MOD << endl;

    return 0;
}