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 
import heapq

n,m=map(int,input().split())
s=[int(x) for x in input().split()]
s.append(-2*(10**9))
s.append(2*(10**9))
kk=dict()

for i in range(0,len(s)):
    kk[s[i]]=1

s.sort()


#lf=-2*(10**9)
#rg=2*(10**9)

q=[]
heapq.heapify(q)

for i in range(1,len(s)-1):
    d1=s[i]-s[i-1]-1
    d2=s[i+1]-s[i]-1
    lf_cap=d1//2
    rg_cap=d2//2
    if(d1%2!=0):
        lf_cap+=1

    t2=0
    t3=0
    if(rg_cap==0):
        t3=10**9
    if(lf_cap==0):
        t2=10**9

    heapq.heappush(q,[min(t3,t2)+1,s[i],t2,t3,lf_cap,rg_cap])

#print(s)
#print(q)

ans=[]
tot=0
while(m>0):

    t=heapq.heappop(q)
    #print('tttttttttttt',t)
    if(t[2]+1==t[0] and m>0):
        ans.append(t[1]-t[0])
        t[2]+=1
        tot+=t[0]
        m-=1
    if(t[3]+1==t[0] and m>0):
        ans.append(t[1]+t[0])
        t[3]+=1
        tot+=t[0]
        m-=1
    
    if(t[5]==t[3]):
        t[3]=10**9
    if(t[4]==t[2]):
        t[2]=10**9

    t[0]=min(t[2],t[3])+1
    heapq.heappush(q,t)
    #print('ansssssssssssssss',ans,t)

print(tot)
print(*ans)
    

    

    
    

    

    
    
    





