You are given an integer m, and a list of n distinct integers between 0 and m - 1.

You would like to construct a sequence satisfying the properties:

  * Each element is an integer between 0 and m - 1, inclusive. 
  * All prefix products of the sequence modulo m are distinct. 
  * No prefix product modulo m appears as an element of the input list. 
  * The length of the sequence is maximized. 



Construct any sequence satisfying the properties above.

Input

The first line of input contains two integers n and m (0 ≤ n < m ≤ 200 000) — the number of forbidden prefix products and the modulus.

If n is non-zero, the next line of input contains n distinct integers between 0 and m - 1, the forbidden prefix products. If n is zero, this line doesn't exist.

Output

On the first line, print the number k, denoting the length of your sequence.

On the second line, print k space separated integers, denoting your sequence.

Examples

Input

0 5


Output

5
1 2 4 3 0


Input

3 10
2 9 1


Output

6
3 9 2 9 8 0

Note

For the first case, the prefix products of this sequence modulo m are [1, 2, 3, 4, 0].

For the second case, the prefix products of this sequence modulo m are [3, 7, 4, 6, 8, 0].
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, m, w[N], vis[N], dp[N], pre[N], cnt;
vector<int> g[N];
long long exgcd(long long &x, long long &y, long long a, long long b) {
  if (!b) {
    x = 1, y = 0;
    return a;
  }
  long long g = exgcd(x, y, b, a % b);
  long long t = x;
  x = y;
  y = t - a / b * y;
  return g;
}
long long getans(long long a, long long b) {
  long long x, y;
  long long c = exgcd(x, y, a, m);
  x = (x * b / c % m + m) % m;
  return x;
}
int main() {
  long long x, y;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    vis[x] = 1;
  }
  for (int i = 0; i < m; i++)
    if (!vis[i]) g[exgcd(x, y, (long long)i, (long long)m)].push_back(i);
  for (int i = 1; i <= m; i++) {
    if (m % i) continue;
    dp[i] += g[i].size();
    for (int j = i + i; j <= m; j += i) {
      if (m % j) continue;
      if (dp[j] <= dp[i]) {
        dp[j] = dp[i];
        pre[j] = i;
      }
    }
  }
  int now = m;
  while (1) {
    for (int i = 0; i < (int)g[now].size(); i++) w[cnt++] = g[now][i];
    if (now == 1) break;
    now = pre[now];
  }
  printf("%d\n", cnt);
  printf("%d ", w[cnt - 1]);
  for (int i = cnt - 1; i; i--) printf("%lld ", getans(w[i], w[i - 1]));
  return 0;
}
