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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>

#define ll long long

using namespace std;

const ll mod = 1e9+7;
ll ara[100005],dp[100005];

///dp[i] - number of ways such that we used i candies so far

void addSelf(ll &x,ll y)
{
    x += y;
    if(x >= mod)
        x -= mod;
}

void subSelf(ll &x,ll y)
{
    x -= y;
    if(x < 0)
        x += mod;
}

void naiveDP(ll n,ll k)
{
    dp[0] = 1;
    for(ll i=1 ; i<=n ; i++){
        for(ll used = k ; used >= 0 ; used--){
            ll lft = used + 1;
            ll rgt = used + min(ara[i] , k-used);

            for(ll j=lft ; j<=rgt ; j++)
                addSelf(dp[j] , dp[used]);
        }
    }
    cout<<dp[k]<<endl;
}

void optimizedDP(ll n,ll k)
{
    dp[0] = 1;
    for(ll i=1 ; i<=n ; i++){
        vector<ll>temp(k+1);
        for(ll used=k ; used >= 0 ; used--){
            ll lft = used + 1;
            ll rgt = used + min(ara[i] , k-used);

            if(lft <= rgt){
                addSelf(temp[lft] , dp[used]);
                if(rgt+1 <= k)
                    subSelf(temp[rgt+1] , dp[used]);
            }
        }
        ll prefSum = 0;
        for(ll x=0 ; x <= k ; x++){
            addSelf(prefSum , temp[x]);
            addSelf(dp[x] , prefSum);
        }
    }
    cout<<dp[k]<<endl;
}

int main()
{
    ll i,j,n,k,ans;
    cin>>n>>k;
    for(i=1 ; i<=n ; i++)   cin>>ara[i];

    optimizedDP(n , k);

    return 0;
}