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)

#include<stdio.h>
#define mod 998244353
#define ll long long
int main(){
    ll n, s, i, j, ans = 0;
    scanf("%lld %lld" ,&n ,&s);
    ll a[n], dp[3000][3001];
    for(i = 0; i < n; i++)
        scanf("%lld" ,&a[i]);
    for(i = 0; i <= s; i++){
        dp[0][i] = 0;
        if(a[0] == i)
            dp[0][i] = 1;
    }
    for(i = 0; i < n; i++)
        dp[i][0] = i + 2;
    for(i = 1; i < n; i++){
        for(j = 1; j < a[i]; j++)
            dp[i][j] = dp[i - 1][j];
        for(j = a[i]; j <= s; j++)
            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - a[i]]) % mod;
    }
    for(i = 0; i < n; i++)
        ans = (ans + dp[i][s]) % mod;
    printf("%lld\n",ans);
    return 0;
}