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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma warning(disable : 4996)
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>
#include<bitset>
#include<limits.h>
#include<cassert>
#include<iterator>
#include<complex>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<time.h>
#include<random>
#include<array>


using namespace std;

#define REP(i, n) for(int i = 0;i < n;++i)
#define REPR(i, n) for(int i = n-1;i >= 0;--i)
#define FOR(i, m, n) for(int i = m;i < n;++i)
#define FORR(i, m, n) for(int i = m-1;i >= n;--i)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout<<(x)<<'\n'
#define pe(x) cout<<(x)<<" "
#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define lb(v,n) lower_bound(v.begin(), v.end(), (n))
#define ub(v,n) upper_bound(v.begin(), v.end(), (n))
#define int long long
//#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i,v.size())cout << v[i] << ((i == v.size() - 1) ? "\n" : " ")
template<typename T1, typename T2> inline void chmin(T1 & a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
typedef pair<int, int>P;
typedef array<int, 3> arr3;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353; constexpr int MAX = 2000020;
const double pi = acos(-1); constexpr double EPS = 1e-8;
constexpr ll INF = 1e18;
long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll add(ll x, ll y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
ll sub(ll x, ll y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
ll mult(ll x, ll y) {
	return (x * y) % MOD;
}
ll bin_pow(ll x, ll p) {
	if (p == 0) return 1;
	if (p & 1) return mult(x, bin_pow(x, p - 1));
	return bin_pow(mult(x, x), p / 2);
}

ll dp[3030][3030];
ll sum[3030][3030];//[0,i)でsumを作る方法
ll sumR[3030][3030];//[i,N)でsumを作る方法
int N, S;
int A[3030];


void solve() {
	cin >> N >> S;
	REP(i, N)cin >> A[i];
	sum[0][0] = 1;
	ll ans = 0;
	REP(i, N) {
		sum[i][0] = i + 1;
		//sum[i + 1][A[i]] = add(sum[i][A[i]], i + 1);
		FOR(s,0, S + 1) {
			if(s>0)sum[i + 1][s] += sum[i][s];
			if (s + A[i] <= S) {
				sum[i + 1][s + A[i]] += sum[i][s];
				sum[i + 1][s + A[i]] %= MOD;
			}
		}
		ans += sum[i + 1][S];
		ans %= MOD;
	}
	//REP(i, N + 1) {
	//	REP(j, S + 1) {
	//		pe(i); pe(j); print(sum[i][j]);
	//	}
	//}
	print(ans);
}


signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//int q; cin >> q;
	//while (q--)
	solve();
}