Well task should be easy i guess but i failed to find the answer.

I need to find some regex in file and print all lines from first match till the end of file but the first found one.

To find and print sed/awk do the trick:

```
awk '/regex/,0'
sed '/regex/,$p'

```

But i feel pretty silly that i can't exclude first matching line from output.

Any tips? Pretty sure that shouldn't be hard. I'm just lacking of knowledge.

PS I need exactly to:

* find line matching regex
* print all lines from this regex to the end of file
* exclude first line

in one action.

Excluding line after only find-n-print action leads to more complicated solution

Example
-------

```
$ cat file
11 1st Cookie
12 2nd Cookie
13 1st Dinner
14 1st Candy
15 2nd Candy
16 3rd Cookie
17 1st Cake
18 2nd Dinner
19 2nd Cake

```

Find and print:

```
$ awk '/Dinner/,0' file
13 1st Dinner
14 1st Candy
15 2nd Candy
16 3rd Cookie
17 1st Cake
18 2nd Dinner
19 2nd Cake

```

What i want to get:

```
14 1st Candy
15 2nd Candy
16 3rd Cookie
17 1st Cake
18 2nd Dinner
19 2nd Cake

```