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A. Points in Segments

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputYou 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 $$$r_i$$$ ($$$1 \le l_i \le r_i \le m$$$) — 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 $$$l \le x \le r$$$.

Input

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

The next $$$n$$$ lines contain two integers each $$$l_i$$$ and $$$r_i$$$ ($$$1 \le l_i \le r_i \le m$$$) — 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.

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

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