Problem:
Score : 200 points

Problem Statement

The development of algae in a pond is as follows.

Let the total weight of the algae at the beginning of the year i be x_i gram. For i≥2000, the following formula holds:

x_{i+1} = rx_i - D

You are given r, D and x_{2000}. Calculate x_{2001}, ..., x_{2010} and print them in order.

Constraints

2 ≤ r ≤ 5

1 ≤ D ≤ 100

D < x_{2000} ≤ 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

r D x_{2000}

Output

Print 10 lines. The i-th line (1 ≤ i ≤ 10) should contain x_{2000+i} as an integer.

Sample Input 1

2 10 20

Sample Output 1

30
50
90
170
330
650
1290
2570
5130
10250

For example, x_{2001} = rx_{2000} - D = 2 \times 20 - 10 = 30 and x_{2002} = rx_{2001} - D = 2 \times 30 - 10 = 50.

Sample Input 2

4 40 60

Sample Output 2

200
760
3000
11960
47800
191160
764600
3058360
12233400
48933560

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//保留所有权利。
#include<iostream>
#include<stdio.h>//printするよ
#include<vector>
#include<algorithm>//辞書順はnext_permutationだよ
#include<set>
#include<string>
#include<string.h>
#include<time.h>
#include<random>
#include<cmath>
#include<math.h>
#include<queue>
#include<time.h>
#include<iomanip>
#include<utility>//swapで数値交換するよ
//END OF LIBRARIES.
#define rt "\n"//改行を最速化させるよ
#define rep(i,n) for(int i=0;i<n;i++)
#define rop(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n-1;0<=i;i--)
#define yes(ans) if(ans)cout<<"yes"<<rt;else cout<<"no"<<rt;
#define Yes(ans) if(ans)cout<<"Yes"<<rt;else cout<<"No"<<rt;
#define YES(ans) if(ans)cout<<"YES"<<rt;else cout<<"NO"<<rt;
#define vcin(v) for (int i = 0; i < (v).size(); i++) { cin >> (v)[i]; }
#define sort(s) sort(s.begin(),s.end())
#define reve(s) reverse(s.begin(),s.end())
#define asas int ans=0
#define cncn int cnt=0
#define smsm int sum=0
#define str srting
#define nn int n=0
#define please return
#define AC 0//おまじないだよ
#define Rapid cin.tie(0);ios::sync_with_stdio(false)
//END OF DEFINE.
using namespace std;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vcha;
typedef long long int llint;
typedef pair<int, int> pint;
typedef pair<llint, llint> pllint;
typedef vector<llint> vllint;
typedef vector<pint> vpint;
typedef vector<pair<llint, llint>> vpllint;
typedef vector<vector<int>> vvint;
typedef vector<vector<llint>> vvllint;
typedef vector<vector<pint>> vvpint;
typedef vector<bool> vbool; 
//END OF TYPEDEF.

//最大公約数（GCD）を求めるよ
//最小公倍数（LCM）は＜＜　A × B ÷ GCD　 ＞＞で求まるよ

long long GCD(long long a, long long b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}
long long LCM(long long a, long long b) {
	return a * b / GCD(a, b);
}
int x = 0, y = 0;
//BEGIN OF CODE.
int main(void){
	int r, d;
	cin >> r >> d;
	vllint x(11,0);
	cin >> x[0];
	rep(i, 10) {
		x[i + 1] = x[i] * r - d;
		cout << x[i+1] << rt;
	}
	please AC;
}
//THE END.