There are n Christmas trees on an infinite number line. The i-th tree grows at the position x_i. All x_i are guaranteed to be distinct.

Each integer point can be either occupied by the Christmas tree, by the human or not occupied at all. Non-integer points cannot be occupied by anything.

There are m people who want to celebrate Christmas. Let y_1, y_2, ..., y_m be the positions of people (note that all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct and all y_j should be integer). You want to find such an arrangement of people that the value ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| is the minimum possible (in other words, the sum of distances to the nearest Christmas tree for all people should be minimized).

In other words, let d_j be the distance from the j-th human to the nearest Christmas tree (d_j = min_{i=1}^{n} |y_j - x_i|). Then you need to choose such positions y_1, y_2, ..., y_m that ∑_{j=1}^{m} d_j is the minimum possible.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of Christmas trees and the number of people.

The second line of the input contains n integers x_1, x_2, ..., x_n (-10^9 ≤ x_i ≤ 10^9), where x_i is the position of the i-th Christmas tree. It is guaranteed that all x_i are distinct.

Output

In the first line print one integer res — the minimum possible value of ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| (in other words, the sum of distances to the nearest Christmas tree for all people).

In the second line print m integers y_1, y_2, ..., y_m (-2 ⋅ 10^9 ≤ y_j ≤ 2 ⋅ 10^9), where y_j is the position of the j-th human. All y_j should be distinct and all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct.

If there are multiple answers, print any of them.

Examples

Input


2 6
1 5


Output


8
-1 2 6 4 0 3 


Input


3 5
0 3 1


Output


7
5 -2 4 -1 2 
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long fn(long long x, long long rn[]) {
  if (x == rn[x])
    return x;
  else
    return rn[x] = fn(rn[x], rn);
}
bool un(long long x, long long y, long long rn[], long long sz[]) {
  x = fn(x, rn);
  y = fn(y, rn);
  if (x == y) return false;
  if (sz[x] < sz[y]) swap(x, y);
  sz[x] += sz[y];
  rn[y] = x;
  return true;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long MOD = 998244353;
long long mod = 1000000007;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long inv(long long val, long long MODx = MOD) {
  return power(val, MODx - 2, MODx);
}
vector<long long> fac, ifac;
void preFac(long long sz) {
  fac.resize(sz + 1), ifac.resize(sz + 1);
  fac[0] = 1;
  for (int i = 1; i <= sz; i++) {
    fac[i] = (i * fac[i - 1]) % MOD;
  }
  ifac[sz] = inv(fac[sz]);
  for (int i = sz - 1; i >= 0; i--) {
    ifac[i] = ((i + 1) * ifac[i + 1]) % MOD;
  }
}
long long nCr(long long N, long long R) {
  if (R <= N and R >= 0) {
    return ((fac[N] * ifac[R]) % MOD * ifac[N - R]) % MOD;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long a[n];
  map<long long, long long> vis;
  set<pair<long long, long long>> st;
  for (long long i = 0; i <= n - 1; i++) {
    cin >> a[i];
    vis[a[i]] = 1;
    st.insert({1, a[i] - 1});
    st.insert({1, a[i] + 1});
  }
  long long ans = 0;
  vector<long long> vec;
  while (!st.empty() and m > 0) {
    while (!st.empty()) {
      pair<long long, long long> p = *st.begin();
      if (vis[p.second] == 1) {
        st.erase(st.begin());
      } else {
        break;
      }
    }
    pair<long long, long long> p = *st.begin();
    st.erase(st.begin());
    ans += p.first;
    vec.push_back(p.second);
    vis[p.second] = 1;
    m--;
    st.insert({p.first + 1, p.second - 1});
    st.insert({p.first + 1, p.second + 1});
  }
  cout << ans << "\n";
  for (auto u : vec) {
    cout << u << " ";
  }
}
