Food Buying（CF-1296B）

Problem Description

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

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

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

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

For example, if Mishka has s=19 burles then the maximum number of burles he can spend is 21. Firstly, he can spend x=10 burles, obtain 1 burle as a cashback. Now he has s=10 burles, so can spend x=10 burles, obtain 1 burle as a cashback and spend it too.

You have to answer t independent test cases.

Input

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

The next t lines describe test cases. Each test case is given on a separate line and consists of one integer s (1≤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.

Examples

Input

6
1
10
19
9876
12345
1000000000

Output

1
11
21
10973
13716
1111111111

题意：t 组数据，每组给出 1 个数 n，代表有 n 元钱，现在用这 n 元钱买东西，每买 10 元返现 1 元，不足 10 元不进行返现，在返现后，将剩下的钱与返现的钱相加，继续买东西，重复返现过程，问最终能买多少钱的东西

思路：照题意模拟即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define Pair pair<int,int>
LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; }
LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;}
LL quickMultPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;}
LL quickPowMod(LL a,LL b,LL mod){ LL res=1; while(b){if(b&1)res=(a*res)%mod; a=(a*a)%mod; b>>=1; } return res; }
LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); }
LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); }
LL LCM(LL x,LL y){ return x/GCD(x,y)*y; }
const double EPS = 1E-15;
const int MOD = 1000000000+7;
const int N = 2000+5;
const int dx[] = {0,0,1,-1,1,1,-1,-1};
const int dy[] = {1,-1,0,0,1,-1,1,-1};
using namespace std;

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        LL n;
        scanf("%lld", &n);

        LL res = 0;
        while (n >= 0) {
            LL k = floor(n / 10);
            res += k * 10;
            n -= k * 10;
            n += k;

            if (n <= 9) {
                res += n;
                break;
            }
        }
        printf("%lld\n", res);
    }

}