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)

// HTTF.cpp : このファイルには 'main' 関数が含まれています。プログラム実行の開始と終了がそこで行われます。
//

#include "bits/stdc++.h"
#define YES "YES"
#define NO "NO"
#define YESNO OUT(three(solve(),YES,NO))
#define ECHO OUT(solve())
#define three(A,B,C) ((A)?(B):(C))
#define FOR(i,a,b)  for(LL i=(a);i< (LL)(b);i++)
#define EFOR(i,a,b) for(LL i=(a);i<=(LL)(b);i++)
#define RFOR(i,a,b) for(LL i=(a);i>=(LL)(b);i--)
#define REP(i,b) FOR(i,zero,b)
#define EREP(i,b) EFOR(i,zero,b)
#define RREP(i,b) RFOR(i,b,zero)
#define ALL(c) c.begin(),c.end()
#define UNIQUE(c) sort(ALL(c));c.erase(unique(ALL(c)),c.end())
#define MAX(c) (*max_element(ALL(c)))
#define MIN(c) (*min_element(ALL(c)))
#define MP make_pair
#define FI first
#define SE second
#define SI(x) (LL(x.size()))
#define PB push_back
#define DEBUG(a) OUT(a)
#define DEBUG2(a,b) OUT2(a,b)
#define cat cout << __LINE__ << endl
#define OUT(a) cout << (a) << endl
#define OUT2(a,b) cout << (a) <<" "<<(b) << endl
#define zero 0LL
#define int LL
#define pb emplace_back
#define eb pb

using namespace std;
template<typename T> inline void maximize(T &a, T b) { a = max(a, b); }
template<typename T> inline void minimize(T &a, T b) { a = min(a, b); }
template<typename T> inline bool middle(T a, T b, T c) { return b <= a && a <= c; }
template<class T> inline bool MX(T &l, const T &r) { return l < r ? l = r, 1 : 0; }
template<class T> inline bool MN(T &l, const T &r) { return l > r ? l = r, 1 : 0; }
typedef long long LL;
typedef double ld;
typedef int ut;
typedef vector<ut> VI;
typedef vector<VI> VII;
typedef pair<ut, ut> pr;
typedef pair<ut, pr> ppr;
typedef vector<pr> Vpr;
typedef vector<ppr> Vppr;
typedef priority_queue<pr, Vpr, greater<pr> > PQ;
inline void outputVI(VI x) { REP(i, SI(x)) { cout << three(i, " ", "") << x[i]; }OUT(""); }
const int SIZE1 = 5e5 + 1000;
const int SIZE2 = 3010;
const int SIZE3 = 330;
const int SIZE = SIZE1;
const int MAPSIZE = 40;
const LL p = 7 + 1e9;
const LL INF = 1LL << 60;
const long double EPS = 1e-7;
typedef pair<ld, ut> pld;

#define endl "\n" //インタラクティブでは消す
ut N, M, K, L, Q, D, H, W;
// ut A,B,C,D,E,F,G,H,I,J,O,P,Q,R,T,U;
VI edges[SIZE];
LL vals[SIZE], maps2[SIZE2][SIZE2], answer = zero;
LL maps[SIZE2][SIZE2];
LL DP[101][SIZE];
LL DP2[101][SIZE];
bool finished[101][SIZE];
bool finished2[101][SIZE];
LL A[SIZE];
string s, t;
LL solve2(LL, LL);
LL sum(LL n, LL x) {
	if (x < 0) return 0;
	if (finished2[n][x]) return DP2[n][x];
	finished2[n][x] = true;
	return DP2[n][x] = (sum(n, x - 1) + solve2(n, x))%p;
}
LL sum(LL n, LL a, LL b) {
	return (sum(n, b) - sum(n, a - 1)+p)%p;
}
LL solve2(LL n,LL k) {
	if (k < 0) return 0;
	if (n==0) return k==0;
	if (finished[n][k]) return DP[n][k];
	finished[n][k] = true;
	DP[n][k] = sum(n - 1, k - A[n], k);
//	cout << n << " " << k << " " << DP[n][k] << endl;
	return DP[n][k]%p;
}
LL solve() {
	cin >> N >>K;
	EFOR(i,1, N) cin >> A[i];
	REP(i, N) REP(j, K) sum(i, j);
	cout << solve2(N,K) << endl;
	return 0;
}




//!!!!!!!!!!!!!!!!!!!実装を詰める!!!!!!!!!!!!!!!!!!!!!!!!!
signed main() {

	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(12);
	solve();
	cin >> N;
	return 0;

}