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 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <numeric>
#include <list>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <tuple>
#include <deque>
#include <complex>

using namespace std;

/*
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
*/

typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<long long, long long> pll;
typedef vector<pll> vpll;

typedef long double ld;
typedef vector<ld> vld;

typedef vector<bool> vb;

#define rep(i, n) for (ll i = 0; i < (n); i++)
#define reps(i, n) for (ll i = 1; i <= (n); i++)
#define rrep(i, n) for (ll i = (n) - 1; i >= 0; i--)
#define rreps(i, n) for (ll i = (n); i >= 1; i--)
#define all(v) (v).begin(), (v).end()

template <class T> void chmin(T& a, T b) { a = min(a, b);}
template <class T> void chmax(T& a, T b) { a = max(a, b);}

constexpr int INF = 1 << 30;
constexpr ll INFL = 1LL << 60;
constexpr ll MOD = 998244353;
constexpr ld EPS = 1e-12;
ld PI = acos(-1.0);

struct mint {
	ll x;
	mint(ll x=0):x((x%MOD+MOD)%MOD){}
	mint operator-() const { return mint(-x);}
	mint& operator+=(const mint a) {if ((x+=a.x)>=MOD) x-=MOD; return *this;}
	mint& operator-=(const mint a) {if ((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
	mint& operator*=(const mint a) {(x*=a.x)%=MOD; return *this;}
	mint operator+(const mint a) const {mint res(*this); return res+=a;}
	mint operator-(const mint a) const {mint res(*this); return res-=a;}
	mint operator*(const mint a) const {mint res(*this); return res*=a;}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t>>1);
		a *= a;
		if (t&1) a *= *this;
 		return a;
	}

	// For prime mod.
	// Do not use if MOD is not prime number !!
	mint inv() const { return pow(MOD-2);}
	mint& operator/=(const mint a) { return (*this) *= a.inv();}
	mint operator/(const mint a) const {mint res(*this); return res/=a;}
};

typedef vector<mint> vm;
typedef vector<vm> vvm;

void solve() {
    ll n, s;
    cin >> n >> s;
    vll a(n);
    for(ll i = 0; i < n; i++) cin >> a[i];
	vvm dp(n + 1, vm(s + 1));
	dp[0][0] = 1;
	for(ll i = 0; i < n; i++) {
		dp[i + 1][0] += 1;
		for(ll j = 0; j <= s; j++) {
			dp[i + 1][j] += dp[i][j];
			if(j + a[i] <= s) dp[i + 1][j + a[i]] += dp[i][j];
		}
	}
	mint ans;
	for(ll i = 0; i <= n; i++) ans += dp[i][s];
	cout << ans.x << endl;
}

int main() {
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    solve();
}