/*
扩展欧几里得算法
给予二整数 a 与 b, 必存在有整数 x 与 y 使得ax + by = gcd(a,b)
设aX' + bY' = gcd(a, b)
bX'' + (a mod b)Y'' = gcd(b, a mod b)
bX'' + (a mod b)Y'' = gcd(a, b)
bX'' + (a mod b)Y'' = aX' + bY'
bX'' + [a - b * (a / b)]Y'' = aX' + bY'
bX'' + aY'' - b * (a / b)Y'' = aX' + bY'
aY'' + b[X'' - Y'' * (a / b)] = aX' + b Y'
X' = Y''
Y' = X'' -Y'' * (a / b)
*/
public class Main {
public static void main(String[] args) {
System.out.println();
}
static int x, y;
static int extgcd(int a, int b) {
int d = a;
if (b == 0) {
x = 1;
y = 0;
} else {
d = extgcd(b, a % b);
int t = x;
x = y;
y = t - y * (a / b);
}
return d;
}
}