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 <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
//using namespace __gnu_pbds;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair <int,int> pii;
typedef pair <ll,ll> pll;

#define si(a)           scanf("%d",&a)
#define sii(a,b)        scanf("%d %d",&a,&b)
#define siii(a,b,c)     scanf("%d %d %d",&a,&b,&c)

#define sl(a)           scanf("%lld",&a)
#define sll(a,b)        scanf("%lld %lld",&a,&b)
#define slll(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)

#define un(x)           x.erase(unique(all(x)), x.end())
#define xx              first
#define yy              second
#define pb              push_back
#define mp              make_pair
#define all(v)          v.begin(),v.end()
#define D(x)            cerr << #x " = " << x << '\n'
#define DBG             cerr << "Hi!" << '\n'

#define CLR(a)          memset(a,0,sizeof(a))
#define SET(a)          memset(a,-1,sizeof(a))

#define PI              acos(-1.0)

inline int setBit(int n,int pos) { return n = n | (1 << pos); }
inline int resetBit(int n,int pos) { return n = n & ~(1 << pos); }
inline bool checkBit(int n,int pos) { return (bool)(n & (1 << pos)); }
//inline int countBit(ll n) { return __builtin_popcountll(n); }


//int fx[] = {+0, +0, +1, -1, -1, +1, -1, +1};
//int fy[] = {-1, +1, +0, +0, +1, +1, -1, -1}; //Four & Eight Direction

/******************************************************************************************/

//const int MAX = 200010;
//const int INF = 0x3f3f3f3f;
//const double inf = 1.0/0.0;
const int MOD = 1000000007;

int ara[105];
int dp[105][100005];
int cum[100005];

int getcum(int i,int j) {
    if(i<0) i=0;
    if(i==0) return cum[j];
    else return (cum[j]-cum[i-1] + MOD) % MOD;
}

int main() {
//    freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int n,k;
    sii(n,k);
    for(int i=1;i<=n;i++) si(ara[i]);
    dp[0][0] = cum[0] = 1;
    for(int j=1;j<=k;j++) cum[j] += cum[j-1];
    for(int i=1;i<=n;i++) {
        dp[i][0] = 1;
        for(int j=1;j<=k;j++) {
            dp[i][j] = getcum(j-ara[i],j);
        }
        cum[0] = 1;
        for(int j=1;j<=k;j++) {
            cum[j] = (cum[j-1] + dp[i][j]) % MOD;
        }
    }
    cout << dp[n][k] << endl;
    return 0;
}