FZU 2278 YYSYinyangshi is a famous RPG game on mobile phones

Yinyangshi is a famous RPG game on mobile phones.

Kim enjoys collecting cards in this game. Suppose there are n kinds of cards. If you want to get a new card, you need to pay W coins to draw a card. Each time you can only draw one card, all the cards appear randomly with same probability 1/n. Kim can get 1 coin each day. Suppose Kim has 0 coin and no cards on day 0. Every W days, Kim can draw a card with W coins. In this problem ,we define W=(n-1)!.

Now Kim wants to know the expected days he can collect all the n kinds of cards.

Input

The first line an integer T(1 ≤ T ≤ 10). There are T test cases.

The next T lines, each line an integer n. (1≤n≤3000)

Output

For each n, output the expected days to collect all the n kinds of cards, rounded to one decimal place.

Sample Input

```
4
1
2
5
9
```

Sample Output

十几次失败之后终于过了，数要4位存一次，一开始用结构体一直tle。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mx = 9200;
const int mod = 1e4;
int sum[mx], jie[mx], te[mx]; 
int po;
void dis(){
    printf("%d",sum[po]);
	for(int i = po - 1; i>=0; i--)
		printf("%04d", sum[i]);
	puts(".0");

}

void mul( int x){

	for(int i = 0; i <= po; i++){
		jie[i] = jie[i] * x;
	
	}
	
	for(int i = 0; i <= po; i++){
		if(jie[i] >= mod){
		    jie[i + 1] += jie[i] / mod;
			jie[i] = jie[i] % mod;	
		}
		  
	}
	if(jie[po + 1] > 0)
	   po++;	
}
void add(int x){
	int cnt = 0;
	for(int i = po; i >= 0; i--){
		cnt = cnt * mod + jie[i];
		if(cnt < x){
		
			te[i] = 0;
			continue;
		} 
	  te[i] = cnt / x;
	  cnt = cnt % x;
	}
	
	for(int i = po; i >= 0; i--){
      sum[i]+=te[i];	
	}
	
	for(int i = 0; i <= po; i++){
      	if(sum[i] >= mod){
		    sum[i + 1] += sum[i] / mod;
			sum[i] = sum[i] % mod;	
		}	
	}
	
	if(sum[po + 1] > 0)
	   po++;
	
}
int main(){
	int T, n, x;
	scanf("%d", &T);
	//T = 90;
	while(T--){
		scanf("%d", &n);
		memset(sum, 0, sizeof(sum));
		memset(jie, 0, sizeof(jie));
		jie[0] = 1; po = 0;
		for(int i = 2; i <= n; i++){
			mul(i);
		}
		
		for(int i = 1; i <= n; i++){
			add(i);
		}
			
	dis();

}	
	return 0;
}

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