Problem:
Score : 600 points

Problem Statement

Given are a sequence of N integers A_1, A_2, \ldots, A_N and a positive integer S.

For a pair of integers (L, R) such that 1\leq L \leq R \leq N, let us define f(L, R) as follows:

f(L, R) is the number of sequences of integers (x_1, x_2, \ldots , x_k) such that L \leq x_1 < x_2 < \cdots < x_k \leq R and A_{x_1}+A_{x_2}+\cdots +A_{x_k} = S.

Find the sum of f(L, R) over all pairs of integers (L, R) such that 1\leq L \leq R\leq N. Since this sum can be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N \leq 3000

1 \leq S \leq 3000

1 \leq A_i \leq 3000

Input

Input is given from Standard Input in the following format:

N S
A_1 A_2 ... A_N

Output

Print the sum of f(L, R), modulo 998244353.

Sample Input 1

3 4
2 2 4

Sample Output 1

5

The value of f(L, R) for each pair is as follows, for a total of 5.

f(1,1) = 0

f(1,2) = 1 (for the sequence (1, 2))

f(1,3) = 2 (for (1, 2) and (3))

f(2,2) = 0

f(2,3) = 1 (for (3))

f(3,3) = 1 (for (3))

Sample Input 2

5 8
9 9 9 9 9

Sample Output 2

0

Sample Input 3

10 10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

152

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
char buf[20000];sz;
inline int rd()
{
	int ret=buf[sz++]-'0';
	if(buf[sz]>='0')
	{
		ret=ret*10+buf[sz++]-'0';
		if(buf[sz]>='0')
		{
			ret=ret*10+buf[sz++]-'0';
			if(buf[sz]>='0')
			{
				ret=ret*10+buf[sz++]-'0';
				if(buf[sz]>='0')
				{
					ret=ret*10+buf[sz++]-'0';
				}
			}
		}
	}
	sz++;
	return ret;
}
int d[3001],d2[3001];
const int mod=998244353;
main()
{
	read(0,buf,20000);
	int N=rd();
	int S=rd();
	int ans=0;
	for(int i=1;i<=N;i++)
	{
		d[S]=i;
		int A=rd();
		for(int j=A;j<=S;j++)d2[j]=d[j];
		for(int j=0;j<=S-A;j++)
		{
			d[j]+=d2[j+A];
			d[j]-=mod*(d[j]>=mod);
		}
		ans+=d[0];
		ans-=mod*(ans>=mod);
	}
	printf("%d",ans);
}