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>
#include <sys/types.h>
#include <unistd.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max(x,0.0L));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

const pid_t pid = getpid();
// Problem Specific Parameter:
template<unsigned MOD>  class ModInt {
public:
	ModInt(): x(0) {}
	ModInt(signed y) : x(y >= 0 ? y % MOD : MOD - (-y) % MOD) {}
	ModInt(signed long long y) : x(y >= 0 ? y % MOD : MOD - (-y) % MOD) {}


	// Arithmetic Oprators
	ModInt &operator+=(ModInt that) {
		if ((x += that.x) >= MOD) x -= MOD; 
		return *this;
	}
	ModInt &operator-=(ModInt that) {
		if ((x += MOD - that.x) >= MOD) x -= MOD; 
		return *this;
	}
	ModInt &operator*=(ModInt that) {
		x = 1LL * x * that.x % MOD;
		return *this;
	}
	ModInt &operator/=(ModInt that) {
		return *this *= ModInt(get<1>(extgcd(that.x, int(MOD))));
	}
	ModInt &operator%=(ModInt that) {
		x %= that.x;
		return *this;
	}

	ModInt &operator+=(const int that) { return *this += ModInt(that);}
	ModInt &operator-=(const int that) { return *this -= ModInt(that);}
	ModInt &operator*=(const int that) { return *this *= ModInt(that);}
	ModInt &operator/=(const int that) { return *this /= ModInt(that);}
	ModInt &operator%=(const int that) { return *this %= ModInt(that);}

	// Comparators
	bool operator <(ModInt that) { return x < that.x; }
	bool operator >(ModInt that) { return x > that.x; }
	bool operator<=(ModInt that) { return x <= that.x; }
	bool operator>=(ModInt that) { return x >= that.x; }
	bool operator!=(ModInt that) { return x != that.x; }
	bool operator==(ModInt that) { return x == that.x; }

	// Utilities
	unsigned getval() const { return x;}
	ModInt operator+(ModInt that) const { return ModInt(*this) += that;}
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that;}
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that;}
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that;}
	ModInt operator%(ModInt that) const { return ModInt(*this) %= that;}
	ModInt operator+(const int that) const { return ModInt(*this) += that;}
	ModInt operator-(const int that) const { return ModInt(*this) -= that;}
	ModInt operator*(const int that) const { return ModInt(*this) *= that;}
	ModInt operator/(const int that) const { return ModInt(*this) /= that;}
	ModInt operator%(const int that) const { return ModInt(*this) %= that;}
	ModInt operator=(const int that) { return *this = ModInt(that);}
	friend istream &operator>>(istream& is, ModInt &that) {ll tmp; is >> tmp; that = ModInt(tmp);return is;}
	friend ostream &operator<<(ostream& os, const ModInt &that) { return os << that.x; }
	
	ModInt power(ll n) const {
		ll b = 1LL, a = x;
		while(n){
			if(n & 1) b = b * a % MOD;
			a = a * a % MOD;
			n >>= 1;
		}
		return ModInt(b);
	}
private:
	unsigned x;
	
	inline auto extgcd(int a, int b) {
		if (b == 0) return make_tuple(a, 1, 0);
		tuple<int, int, int> ret = extgcd(b, a % b);
		swap(get<1>(ret), get<2>(ret));
		get<2>(ret) -= a / b * get<1>(ret);
		return ret;
	}
};

const int mod = 1000000007;
using mint = ModInt<mod>;
const mint ZERO = mint(0);
const mint ONE = mint(1);


const int limit = 100010;
mint dp[110][limit];

int main(void){
	int n,k;
	cin >> n >> k;
	vector<int> a(n);
	rep(i,n) cin >> a[i];

	dp[0][0] = ONE;
	rep(i,1,n+1)rep(j,k+1){
		mint cur = ZERO;
		if(j - 1 >= 0) cur += dp[i][j-1];
		cur += dp[i - 1][j];
		if(j - a[i - 1] - 1 >= 0) cur -= dp[i - 1][j - a[i - 1] - 1];
		dp[i][j] = cur;
	}

	cout << dp[n][k] << endl;

	return 0;
}