CodeForces - 1000D：Yet Another Problem On a Subsequence （DP+组合数）

The sequence of integers [3,7,8],[2,5,4,1],[0] — are not.

A sequence of integers is called good if it can be divided into a positive number of good arrays. Each good array should be a subsegment of sequence and each element of the sequence should belong to exactly one array. For example, the sequences [1,2,3,−3−9,4,1] — are not.

For a given sequence of numbers, count the number of its subsequences that are good sequences, and print the number of such subsequences modulo 998244353.

Input

The first line contains the number a1,a2,…,an (−109≤ai≤109) — the sequence itself.

Output

In the single line output one integer — the number of subsequences of the original sequence that are good sequences, taken modulo 998244353.

Examples

Input

3
2 1 1

Output

Input

4
1 1 1 1

Output

Note

In the first test case, two good subsequences — [a2,a3].

In the second test case, seven good subsequences — [a3,a4].

题意：给定序列，问有多少子序列（不一定连续），满足可以划分为若干个组，给个组的第一个等于区间长度-1；

思路：因为关键在于区间的第一个，我们从后向前考虑，dp[i]表示以i为开头，满足题意的数量；sum[i]表示i后面可能的情况数量。

对于i：还要取a[i]个，我们假设最后一个数在j位置，那么dp[i]+=C(j-i-1,a[i]-1)*(1+sum[j+1]);

复杂度为O(N^2)；

#include<bits/stdc++.h>
#define ll long long 
using namespace std;
const int Mod=998244353;
const int maxn=1010;
int a[maxn],dp[maxn],sum[maxn];
int c[maxn][maxn],ans;
int main()
{
    int N,i,j;
    scanf("%d",&N);
    for(i=0;i<=N;i++) c[i][0]=1,c[i][1]=i,c[i][i]=1;
    for(i=1;i<=N;i++)
     for(j=1;j<=N;j++)
       c[i][j]=(c[i-1][j]+c[i-1][j-1])%Mod;
    for(i=1;i<=N;i++) scanf("%d",&a[i]); 
    for(i=N;i>=1;i--){
        if(a[i]>0&&i+a[i]<=N){
            for(j=i+a[i];j<=N;j++){
                (dp[i]+=(ll)c[j-i-1][a[i]-1]*(1+sum[j+1])%Mod)%=Mod;
            }
        }
        sum[i]=(sum[i+1]+dp[i])%Mod;
    }
    printf("%d\n",sum[1]);
    return 0;
}