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: Wrong Answer)

#include <iostream>
#include <vector>
#include <cassert>
#include <string>
#include <map>
#include <algorithm>
#include <queue>
using namespace std;
constexpr int MOD = 998244353;

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

	vector<vector<int>> dp(s + 3030, vector<int>(3));
	dp[0][0] = 1;
	for (int i = 0; i < n; ++i) {
		vector<vector<int>> next(s + 3030, vector<int>(3));
		for (int j = 0; j <= s; ++j) {
			// 状態維持
			next[j][0] += dp[j][0];
			next[j][1] += dp[j][1]; // 足さない
			next[j + a[i]][1] += dp[j][1]; // 足す
			next[j][2] += dp[j][2];
			// 区間を始める
			next[j][1] += dp[j][0]; // 加算なし
			next[j + a[i]][1] += dp[j][0]; // 加算あり
			// 区間を終える
			next[j][2] += dp[j][1]; // 加算なし
			next[j + a[i]][2] += dp[j][0] + dp[j][1]; // 加算あり

			next[j][0] %= MOD;
			next[j][1] %= MOD;
			next[j][2] %= MOD;
			next[j + a[i]][1] %= MOD;
			next[j + a[i]][2] %= MOD;
		}
		dp = next;
	}
	cout << dp[s][2];
	return 0;
}