Напомена: ово је незванична копија задатака. Као таква, не гарантује се да ће овај сајт бити одржаван, и немојте се изненадити ако са њега задаци одједном нестану.

Little Danica was completely disgusted by the Committee’s decision to cancel one round of qualifiers this year, and she started breaking everything within reach, including numbers that happened to be nearby. Whenever she sees a pair of natural numbers, she throws one of them at the other so that they collide. When two numbers collide, a new natural number is formed.

When natural numbers \(X\) and \(Y\) collide, each digit of one is compared with the corresponding digit of the other: the ones digit is compared to the ones digit, the tens digit with the tens, the hundreds with the hundreds, and so on. The lower of the two digits disappears, and the higher is used to form the resulting number. If the digits are equal, both will be used, and the result will have more digits than \(X\) and \(Y\). If one of the numbers has fewer digits than the other, the “extra” digits from the other one are always used (even if they are zero).

Danica can’t deal with all the numbers she is surrounded by, and asked you for help in colliding them. To do so, you first need to answer a question: which number will be formed when two numbers \(X\) and \(Y\) collide?

Input format

The first line of standard input contains the first natural number \(X\), and the second line contains the second number \(Y\) that participates in the collision, as described in the problem statement.

Output format

Your program should print one line to the standard output: the newly-formed number that is the result of colliding \(X\) and \(Y\).

Sample 1





Sample 2





Sample 3





Sample 4





Sample 5






In the first testcase, numbers \(73\) and \(28\) collide. Both have two digits, and the result will have the greater of the tens digits \(7\) and \(2\) as its tens digit, and the greater of \(3\) and \(8\) as the ones digit. The result is therefore \(78\), since \(7\) is greater than \(2\) and \(8\) is greater than \(3\).

In the second testcase, numbers \(64\) and \(357\) collide. One has three digits, and the other has two, so the hundreds digit will be the \(3\) taken from \(357\). The remaining digits are handled as in the previous case: \(6\) is greater than \(5\), and \(4\) is greater than \(7\), so the result will be \(367\).

In the third testcase, numbers \(234\) and \(135\) collide. The matching digit pairs are \(2\) and \(1\), \(3\) and \(3\), and \(4\) and \(5\). The hundreds and ones digits are the larger ones in the corresponding pairs: \(2\) and \(5\). In the tenths place the two digits are equal and will both be included in the result, which is then \(2335\).

In the fourth testcase, numbers \(99099\) and \(9999\) collide. As one has five digits and the other four, the lower four digit pairs will be compared, and the topmost digit of \(99099\) will be included in the result by default. There are only two relevant cases: a pair of \(9\) and \(0\), that produces a single \(9\) in the result, and a pair of \(9\) and \(9\), which results in \(99\). The result is then \(99999999\), since the two numbers contain a total of eight nines, which will all be included in the result.

Finally, in the fifth testcase, numbers \(2\) and \(100\) collide. The digits \(1\) and \(0\) are simply copied from the longer number, and the ones digit is \(2\), since it is greater than the corresponding \(0\), making the result \(102\).


Testcases are split into five disjoint groups:


The numbers given in the input will not have leading zeroes. Your output should not contain leading zeroes.