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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

const ll MOD = 998244353;

int main()
{
    ios_base::sync_with_stdio(0);
    int n, S;
    cin >> n >> S;
    vector<int> a(n);
    for (int i = 0; i < n; ++i)
        cin >> a[i];
    vector< vector<ll> > dp(n, vector<ll>(S + 1, 0));
    if (a[0] <= S)
        dp[0][a[0]] = 1;
    for (int i = 1; i < n; ++i)
    {
        for (int j = 1; j <= S; ++j)
        {
            if (j - a[i] >= 0)
                dp[i][j] = (dp[i][j] + dp[i - 1][j - a[i]]) % MOD;
            dp[i][j] = (dp[i][j] + dp[i - 1][j]) % MOD;
        }
        if (a[i] <= S)
            dp[i][a[i]] = (dp[i][a[i]] + i + 1) % MOD;
    }
    ll odp = 0;
    for (int i = 0; i < n; ++i)
        odp = (odp + dp[i][S]) % MOD;
    cout << odp;
    return 0;
}