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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
#define zeroend __builtin_ctz
#define ones __builtin_popcount
#define ends "\n"
#define space ' '
typedef long long ll;
typedef pair<int, int> pii;

/*
void solve(int i, int j, string s){
	if(i==0)
		ans.insert(s2.substr(0, j)+s);
	else if(j==0)
		ans.insert(s1.substr(0, i)+s);
	else if(s1[i-1]==s2[j-1])
		solve(i-1, j-1, s1[i-1]+s);
	else{
		if(dp[i-1][j]<=dp[i][j-1])
			solve(i-1, j, s1[i-1]+s);
		if(dp[i-1][j]>=dp[i][j-1])
			solve(i, j-1, s2[j-1]+s);
	}
}*/


/*
	cin >> n;
	for(int i=0; i<n; ++i)
		cin >> a[i];

	for(int i=1; i<n; ++i){
		for(int j=0; j<i; ++j){
			if(a[j]>a[i]&&low[i]<high[j]+1)
				low[i]=high[j]+1;
			else if(a[j]<a[i]&&high[i]<low[j]+1)
				high[i]=low[j]+1;
		}
		ans=max({ans, low[i], high[i]});
	}

	cout << ans << ends;


*/

const int N = 100, N2=1e5, M = 1e9+7;

bool dp[N2+1];
//array<ll, 3> dp;
//set<string> ans;
//vector<int> low(N, 1), high(N, 1);
ll a[N];

ll n, x;

int main(){
	cin >> n >> x;
	for(int i=0; i<n; ++i)
		cin >> a[i];

	for(int i=1; i<=x; ++i){
		for(int j=0; j<n; ++j){
			if(a[j]<=i)
				dp[i]|=!dp[i-a[j]];
		}
	}

	cout << (dp[x]?"First":"Second") << ends;
}