title: awk - remove character in regex
body: <p>I want to remove <code>1</code> with <code>awk</code> from this regex: <code>^1[0-9]{10}$</code> if said regex is found in any field. I've been trying to make it work with <code>sub</code> or <code>substr</code> for a few hours now, I am unable to find the correct logic for this. I already have the solution for sed: <code>s/^1\([0-9]\{10\}\)$/\1/</code>, I need to make this work with <code>awk</code>.</p>
  
  <p>Edit for input and output example. Input:</p>
  
  <pre><code>10987654321
  2310987654321
  1098765432123    
  </code></pre>
  
  <p>(awk twisted and overcomplicated syntax)</p>
  
  <p>Output:</p>
  
  <pre><code>0987654321
  2310987654321
  1098765432123    
  </code></pre>
  
  <p>Basically the leading <code>1</code> needs to be removed only when it's followed by ten digits. The 2nd and 3rd example lines are correct, 2nd has <code>23</code> in front of <code>1</code>, 3rd has a leading <code>1</code> but it's followed by 12 digits instead of ten. That's what the regex specifies.</p>