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: Runtime Error)

/*
      @author: Charan Sai
*/
#include <bits/stdc++.h>
using namespace std;

#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define int long long
#define ppi pair<int, int>
#define endl "\n"
#define nl cout << "\n"
#define deb(x) cout << #x << " " << x << endl;

int n, k;
const int MAX_N = 102, MAX_K = 1e5 + 2;
const int mod = 1e9 + 7;
int arr[MAX_N];
int dp[MAX_N][MAX_K];
int s[MAX_N][MAX_K];
// O(N*K*K) time = O(NK^2) time
// O(NK) space
int Top_Down(int i = n, int candies = k) {
    // Base Case
    if(i == 0) return (candies == 0);
    if(dp[i][candies] != -1) return dp[i][candies]%mod;
    // Choice
    int ways = 0;
    for(int x = 0; x <= arr[i]; ++x) {
        if(candies >= x)
            ways += (Top_Down(i - 1, candies - x)%mod);
    }
    return (dp[i][candies]=ways%mod);
}

int Bottom_Up() {
    // Base
    dp[0][0] = 1;
    for(int j = 1; j <= k; ++j) dp[0][j] = 0;
    for(int i = 1; i <= n; ++i) dp[i][0] = 1;
    // Choice
        
    for(int i = 1; i <= n; ++i) {
        for(int candies = 1; candies <= k; ++candies) {
            int ways = 0;
            for(int x = 0; x <= arr[i]; ++x) {
                if(candies >= x)
                ways += dp[i-1][candies-x]%mod;
            }
            dp[i][candies]=ways%mod;
        }
    }
    return dp[n][k];
}

/*
dp[0][0] = 1;
for(int j = 1; j <= k; ++j) dp[0][j] = 0;
for(int i = 1; i <= n; ++i) dp[i][0] = 1;

for(i = 1; i <= n; ++i)
{
    s[i-1][0] = dp[i-1][0];
    for(int j = 1; j <= k; ++j) {
        //compute s[i][j]
        s[i-1][j] = (dp[i-1][j] + s[i-1][j-1])%mod;
    }

    for(int j = 1; j <= k; ++j) {
       //compute dp[i][j]
       dp[i][j] = s[i-1][j]
       if(j-arr[i]-1 >= 0)
        dp[i][j] = (dp[i][j] - s[i-1][j-arr[i]-1] + mod)%mod;
    }
}
s(i,j) = T(i,0) + T(i,1) + ..... + T(i,j)
T(i,j) = s(i-1,j) - s(i-1,j-a[i]-1)

*/

int Bottom_Up2() {
    dp[0][0] = 1;
    for(int j = 1; j <= k; ++j) dp[0][j] = 0;
    for(int i = 1; i <= n; ++i) dp[i][0] = 1;
    int i, j;
    for(i = 1; i <= n; ++i)
    {
        s[i-1][0] = dp[i-1][0];
        for(j = 1; j <= k; ++j) {
            //compute s[i][j]
            s[i-1][j] = (dp[i-1][j] + s[i-1][j-1])%mod;
        }

        for(j = 1; j <= k; ++j) {
           //compute dp[i][j]
           dp[i][j] = s[i-1][j];
           if(j-arr[i]-1 >= 0)
            dp[i][j] = (dp[i][j] - s[i-1][j-arr[i]-1] + mod)%mod;
        }
    }
    return dp[n][k]%mod;
}

int32_t main() {
    IOS;
    int t; cin >> t;
    while(t--) {
        memset(dp, -1, sizeof(dp));
        cin >> n >> k;
        for(int i = 1; i <= n; ++i) cin >> arr[i];
        cout << Bottom_Up2()%mod << endl;
    }
}