Mishka wants to buy some food in the nearby shop. Initially, he has ss burles on his card.

Mishka can perform the following operation any number of times (possibly, zero): choose some positive integer number 1≤x≤s1≤x≤s, buy food that costs exactly xx burles and obtain ?x10??x10? burles as a cashback (in other words, Mishka spends xx burles and obtains ?x10??x10? back). The operation ?ab??ab? means aa divided by bb rounded down.

It is guaranteed that you can always buy some food that costs xx for any possible value of xx.

Your task is to say the maximum number of burles Mishka can spend if he buys food optimally.

For example, if Mishka has s=19s=19 burles then the maximum number of burles he can spend is 2121. Firstly, he can spend x=10x=10 burles, obtain 11 burle as a cashback. Now he has s=10s=10 burles, so can spend x=10x=10 burles, obtain 11 burle as a cashback and spend it too.

You have to answer tt independent test cases.

Input
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The next tt lines describe test cases. Each test case is given on a separate line and consists of one integer ss (1≤s≤1091≤s≤109) — the number of burles Mishka initially has.

Output
For each test case print the answer on it — the maximum number of burles Mishka can spend if he buys food optimally.

Example
Input
6
1
10
19
9876
12345
1000000000
Output
1
11
21
10973
13716
1111111111
每次都盡可能大的拿10的倍數個，按照題意模擬就行了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;int n;int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);int ans=0;while(n){int sum=n/10;if(sum==0){ans+=n;break;}ans+=sum*10;n=sum+n%10;}cout<<ans<<endl;}return 0; }

努力加油a啊，(^o)/~

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