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.
def answer():
    a = input().split()
    a = [int(x) for x in a]
    n = a[0]
    high = a[1]
    i=1
    c=[]
    while n:
        a = [int(x) for x in input().split()]
        c.append(a)            
        n-=1
    low=1
    z=[low]
    i=low

    while i<=high:
        z.append(i)
        i+=1
    d=[]
    for x in z:
        if x not in d:
            d.append(x)
    if len(d)==0:
        print(0)
        done=1
    
    else:
        for x in c:
            i=min(x)
            while i<=max(x):
                if i in d: d.remove(i)
                i+=1
    print(len(d))
    c_out=""
    i=0
    while i<len(d):
        c_out+=str(d[i])
        if i!=len(d)-1:
            c_out+=" "
        i+=1
    print(c_out)
answer()
