Sort a concatenated sequence Consider a sequence based on recurrence relations, f(n) = f(n-1)+f(n-2), starting with f(1) = x1, f(2) = x2. For x1 = 2, x2 = 1, the sequence begins like this: 

2  1  3  4  7  11  18  29  47  76  123  199  322  521  843


Concatenating this into a string will give:

213471118294776123199322521843


Now, divide this list into the smallest possible numbers that gives y(n) &gt; y(n-1). Start with the first number, then the second etc. The first output number should always be a single digit. Pad the last number with the required number of zeros. 

2 13 47 111 829 4776 12319 93225 218430


You will get two numbers, (x1, x2) as input, on any convenient format, and the challenge is to output the sorted list. 

Rules:


Function and programs are OK
The initial sequence shall have exactly 15 numbers (The last number is f(15)). 
x1 and x2 are non-negative (zero is possible).
The output can be on any convenient format
The output vector y must be created so that y2 &gt; y1.


First the smallest possible y1, then the smallest possible y2, then y3 and so on.

If x1 = x2 = 0 then output 15 zeros (on the same format as other output, i.e. not 000000000000000).


Examples:

Input: 1 1
Output: 1  12  35  81  321  345  589  1442  3337 7610

Input: 3 2
Output: 3  25  71  219  315  0811 3121  23435 55898 145300
                             |
                             Optional leading zero 
Input: 0 0
Output: 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0


The shortest code in bytes wins. Please include a link to an online interpreter if possible.
JavaScript ES6, 127 135

(a,b)=&gt;eval("for(n=r=[],v=13,o=a+n+b;v--;a=b,b=t)o+=t=b+a;for(d of o+'0'.repeat(99))(n+=d)&gt;+v&amp;&amp;(r.push(v=n),n='');+v?r:[...o]")


Test



F=(a,b)=&gt;eval("for(n=r=[],v=13,o=a+n+b;v--;a=b,b=t)o+=t=b+a;for(d of o+'0'.repeat(99))(n+=d)&gt;+v&amp;&amp;(r.push(v=n),n='');+v?r:[...o]")

// less golfed

U=(a,b)=&gt;{
  for(n=r=[], o=a+n+b, v=13; v--; a=b, b=t)
    o+= t= b+a;
  for(d of o+'0'.repeat(99))
    if ((n+=d) &gt; +v)
      r.push(v=n), n='';
  return +v ? r : [...o]
}

function test(){
  var i = I.value.match(/\d+/g)
  O.textContent = i.length &gt; 1 ? F(+i[0],+i[1]) : ''
}
test()
A,B : &lt;input id=I value='0 1' oninput='test()'&gt;
&lt;pre id=O&gt;&lt;/pre&gt;