likelihood 12, he leaves his present club, then, at that point, makes another club himself and goes along with it. There is just a single understudy (himself) in the new club he makes. With likelihood 12, he doesn't make new clubs. For this situation, he changes his club to another one (perhaps a similar club he is in right now) with likelihood relative to the number of understudies
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With likelihood 12, he leaves his present club, then, at that point, makes another club himself and goes along with it. There is just a single understudy (himself) in the new club he makes.
With likelihood 12, he doesn't make new clubs. For this situation, he changes his club to another one (perhaps a similar club he is in right now) with likelihood relative to the number of understudies in it. Officially, assume there are k clubs and there are bi understudies in the I-th club for 1≤i≤k (before the understudy blows up). He leaves his present club, and afterward joins the I-th club with likelihood canister.
We note that when a club becomes unfilled, understudies won't ever go along with this is on the grounds that any understudy who blows up will get a vacant club together with likelihood 0 as per the above assertion.
Homer ponders the normal number of days until each understudy is in a similar club interestingly.
We can demonstrate that the appropriate response can be addressed as an objective number pq with gcd(p,q)=1. In this way, you are requested to find the worth from pq−1mod998244353. It very well may be shown that qmod998244353≠0 under the given limitations of the issue.
Input
The main line contains an integer m (1≤m≤1000) — the number of clubs at first.
The subsequent line contains m integers a1,a2,… ,am (1≤
Output
Print one integer — the normal number of days until each understudy is in a similar club interestingly, modulo 998244353.
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