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)

#pragma GCC optimize("O3")
#pragma GCC target("avx")
#pragma GCC optimize("unroll-loops")
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define DB double
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define rep(i,n) FOR(i,0,n)
//#define MOD 1000000007
#define MOD 998244353
#define INF 1145141919810364364
#define PI 3.14159265358979
#define ERR 0.00000000000001
#define NUM 200010

void swap(LL* a,LL* b){LL tmp=*a;*a=*b;*b=tmp;}
void YN(int a){if(a==1)printf("Yes\n");if(a==0)printf("No\n");}
LL pwmn(LL a,LL n){LL ans=1;while(ans<a)ans*=n;return ans;}
LL GCD(LL a,LL b){LL c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int lcmpr(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return -1;if(*(LL*)a<*(LL*)b)return 1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int ccmpr(const void* a,const void* b){return *(char*)b-*(char*)a;}
int scmp(const void* a,const void* b){return strcmp((char*)a,(char*)b);}
int scmpr(const void* a,const void* b){return strcmp((char*)b,(char*)a);}
LL mod(LL a,LL m){if(a<0)return a%m+m;else return a%m;}
LL DIV(LL a,LL d){LL m=MOD,x=1,y=0,k;while(m){k=d/m;d-=k*m;swap(&m,&d);x-=k*y;swap(&x,&y);}return mod(a*mod(x,MOD),MOD);}
LL FAC(LL a){LL i,ans=1;FOR(i,1,a+1){ans*=i;if(MOD>0&&ans>MOD)ans%=MOD;}return ans;}
LL POW(LL a,LL n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return ans;}
LL fact[NUM],finv[NUM],inv[NUM];
void comi(){LL i;fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}

int main(){
  LL N,S,A[3010],dp[3010]={0},i,j,a,ans=0;
  scanf("%lld%lld",&N,&S);
  rep(i,N)scanf("%lld",&A[i]);
  dp[0]=1;
  rep(i,N)for(j=S-1;j>=0;j--){
    if(j+A[i]>S)continue;
    a=1;
    if(j==0)a*=i+1;
    if(j+A[i]==S)a*=N-i;
    dp[j+A[i]]+=dp[j]*a%MOD;
    dp[j+A[i]]%=MOD;
  }

  printf("%lld\n",dp[S]);
	return 0;
}