#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int MAXINT = ((1 << 30) - 1) * 2 + 1;int A, B, C;
struct Hashmap //哈希表代替map
{static const int Ha = 999917, maxe = 46340;int E, lnk[Ha], son[maxe + 5], nxt[maxe + 5], w[maxe + 5];int top, stk[maxe + 5];void clear() { E = 0; while (top) lnk[stk[top--]] = 0; }void Add(int x, int y) { son[++E] = y; nxt[E] = lnk[x]; w[E] = MAXINT; lnk[x] = E; }bool count(int y){int x = y%Ha;for (int j = lnk[x]; j; j = nxt[j])if (y == son[j]) return true;return false;}int& operator [] (int y){int x = y%Ha;for (int j = lnk[x]; j; j = nxt[j])if (y == son[j]) return w[j];Add(x, y); stk[++top] = x; return w[E];}
};
Hashmap f;int exgcd(int a, int b, int &x, int &y)
{if (!b) { x = 1; y = 0; return a; }int r = exgcd(b, a%b, x, y), t = x; x = y; y = t - a / b*y;return r;
}
int BSGS(int A, int B, int C)
{if (C == 1) if (!B) return A != 1; else return -1;if (B == 1) if (A) return 0; else return -1;if (A%C == 0) if (!B) return 1; else return -1; //几种特判int m = ceil(sqrt(C)), D = 1, Base = 1; f.clear();for (int i = 0; i <= m - 1; i++) //先把A^j存进哈希表
    {f[Base] = min(f[Base], i);Base = ((LL)Base*A) % C;}for (int i = 0; i <= m - 1; i++){int x, y, r = exgcd(D, C, x, y);x = ((LL)x*B%C + C) % C; //扩欧求A^jif (f.count(x)) return i*m + f[x]; //找到了D = ((LL)D*Base) % C;}return -1;
}
int main()
{while (~scanf("%d%d%d", &C, &A, &B)){int ans = BSGS(A, B, C);if (ans == -1) printf("no solution\n"); elseprintf("%d\n", ans);}return 0;
}