You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — coordinates of the left and of the right endpoints. 

Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l; r] if and only if l ≤ x ≤ r.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of segments and the upper bound for coordinates.

The next n lines contain two integers each l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that l_i=r_i, i.e. a segment can degenerate to a point.

Output

In the first line print one integer k — the number of points that don't belong to any segment.

In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.

If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.

Examples

Input

3 5
2 2
1 2
5 5


Output

2
3 4 


Input

1 7
1 7


Output

0

Note

In the first example the point 1 belongs to the second segment, the point 2 belongs to the first and the second segments and the point 5 belongs to the third segment. The points 3 and 4 do not belong to any segment.

In the second example all the points from 1 to 7 belong to the first segment.
#declaração de matrizes
#i++

s = input();
temp = s.split(" ")

n = int(temp[0])
m = int(temp[1])

u = [[0,0]]*n
v = [0]*m
#u = [[0 for x in range(w)] for y in range(h)] 

i = 0
while i < n:
    s = input()
    #s.split(" ")
    #print("teste s:", s)
    
    temp = s.split(" ")
    #print(i, temp)
    #u[i][0] = int(temp[:][0])
    #u[i][1] = int(temp[:][1])
    u[i] = [int(temp[:][0]), int(temp[:][1])]
    #print("u", u[i][0], u[i][1])
    i += 1

#print(u)
x = 1
i = 0
k = 0

#print("teste u")
while x <= m:
    i = 0
    pertence = False
    while i < n and not pertence:
        #print(x, i, u[i][0], u[i][1])
        if x >= u[i][0] and x <= u[i][1]:
            pertence = True
        i += 1
    if not pertence:
        v[k] = x
        k += 1
    x += 1  #identaçao no while!

print(k)
i = 0
while i < k:
    print(v[i], end=" ")
    i += 1
