#include <bits/stdc++.h>
using namespace std;
const int H = 16;
const int N = 1 << 18;
int p;
int fpow(int x, int y = p - 2) {
  if (!y) return 1;
  long long z = fpow(x, y >> 1);
  z = z * z % p;
  if (y & 1) z = z * x % p;
  return z;
}
int l[2], a[2][N];
void dfs(int k, int u, int t, int e) {
  u = (u + p) % p;
  a[e][l[e]++] = u;
  if (k == H) return;
  k++;
  if (t != 2) dfs(k, u + 1, 1, e);
  if (t != 1) dfs(k, u - 1, 2, e);
  if (t != 3) dfs(k, fpow(u), 3, e);
}
int f[2];
vector<int> v[2];
void getans(int k, int u, int t, int e, int j) {
  u = (u + p) % p;
  if (!f[e]) v[e].push_back(t);
  if (!f[e] && u == j) {
    f[e] = 1;
    return;
  }
  if (k == H) {
    if (!f[e]) v[e].pop_back();
    return;
  }
  k++;
  if (t != 2) getans(k, u + 1, 1, e, j);
  if (t != 1) getans(k, u - 1, 2, e, j);
  if (t != 3) getans(k, fpow(u), 3, e, j);
  if (!f[e]) v[e].pop_back();
}
int main() {
  int i, j, x, y, z;
  cin >> x >> y >> p;
  dfs(0, x, 0, 0), dfs(0, y, 0, 1);
  sort(a[0], a[0] + l[0]);
  for (i = 0; i < l[1]; i = i + 1) {
    j = lower_bound(a[0], a[0] + l[0], a[1][i]) - a[0];
    if (j < l[0] && a[0][j] - a[1][i] <= 160) break;
  }
  z = a[0][j] - a[1][i];
  getans(0, x, 0, 0, a[0][j]), getans(0, y, 0, 1, a[1][i]);
  cout << v[0].size() - 1 + z + v[1].size() - 1 << endl;
  for (i = 1; i < v[0].size(); i = i + 1) cout << v[0][i] << ' ';
  for (i = 0; i < z; i = i + 1) cout << 2 << ' ';
  for (i = v[1].size() - 1; i > 0; i = i - 1) {
    if (v[1][i] < 3) v[1][i] = 3 - v[1][i];
    cout << v[1][i] << ' ';
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following problem:

Allen is playing Number Clicker on his phone.

He starts with an integer u on the screen. Every second, he can press one of 3 buttons.

  1. Turn u → u+1 \pmod{p}. 
  2. Turn u → u+p-1 \pmod{p}. 
  3. Turn u → u^{p-2} \pmod{p}. 



Allen wants to press at most 200 buttons and end up with v on the screen. Help him!

Input

The first line of the input contains 3 positive integers: u, v, p (0 ≤ u, v ≤ p-1, 3 ≤ p ≤ 10^9 + 9). p is guaranteed to be prime.

Output

On the first line, print a single integer ℓ, the number of button presses. On the second line, print integers c_1, ..., c_ℓ, the button presses. For 1 ≤ i ≤ ℓ, 1 ≤ c_i ≤ 3.

We can show that the answer always exists.

Examples

Input

1 3 5


Output

2
1 1


Input

3 2 5


Output

1
3

Note

In the first example the integer on the screen changes as 1 → 2 → 3.

In the second example the integer on the screen changes as 3 → 2. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int H = 16;
const int N = 1 << 18;
int p;
int fpow(int x, int y = p - 2) {
  if (!y) return 1;
  long long z = fpow(x, y >> 1);
  z = z * z % p;
  if (y & 1) z = z * x % p;
  return z;
}
int l[2], a[2][N];
void dfs(int k, int u, int t, int e) {
  u = (u + p) % p;
  a[e][l[e]++] = u;
  if (k == H) return;
  k++;
  if (t != 2) dfs(k, u + 1, 1, e);
  if (t != 1) dfs(k, u - 1, 2, e);
  if (t != 3) dfs(k, fpow(u), 3, e);
}
int f[2];
vector<int> v[2];
void getans(int k, int u, int t, int e, int j) {
  u = (u + p) % p;
  if (!f[e]) v[e].push_back(t);
  if (!f[e] && u == j) {
    f[e] = 1;
    return;
  }
  if (k == H) {
    if (!f[e]) v[e].pop_back();
    return;
  }
  k++;
  if (t != 2) getans(k, u + 1, 1, e, j);
  if (t != 1) getans(k, u - 1, 2, e, j);
  if (t != 3) getans(k, fpow(u), 3, e, j);
  if (!f[e]) v[e].pop_back();
}
int main() {
  int i, j, x, y, z;
  cin >> x >> y >> p;
  dfs(0, x, 0, 0), dfs(0, y, 0, 1);
  sort(a[0], a[0] + l[0]);
  for (i = 0; i < l[1]; i = i + 1) {
    j = lower_bound(a[0], a[0] + l[0], a[1][i]) - a[0];
    if (j < l[0] && a[0][j] - a[1][i] <= 160) break;
  }
  z = a[0][j] - a[1][i];
  getans(0, x, 0, 0, a[0][j]), getans(0, y, 0, 1, a[1][i]);
  cout << v[0].size() - 1 + z + v[1].size() - 1 << endl;
  for (i = 1; i < v[0].size(); i = i + 1) cout << v[0][i] << ' ';
  for (i = 0; i < z; i = i + 1) cout << 2 << ' ';
  for (i = v[1].size() - 1; i > 0; i = i - 1) {
    if (v[1][i] < 3) v[1][i] = 3 - v[1][i];
    cout << v[1][i] << ' ';
  }
  return 0;
}

```