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A. Even But Not Even

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLet's define a number ebne (even but not even) if and only if its sum of digits is divisible by $$$2$$$ but the number itself is not divisible by $$$2$$$. For example, $$$13$$$, $$$1227$$$, $$$185217$$$ are ebne numbers, while $$$12$$$, $$$2$$$, $$$177013$$$, $$$265918$$$ are not. If you're still unsure what ebne numbers are, you can look at the sample notes for more clarification.

You are given a non-negative integer $$$s$$$, consisting of $$$n$$$ digits. You can delete some digits (they are not necessary consecutive/successive) to make the given number ebne. You cannot change the order of the digits, that is, after deleting the digits the remaining digits collapse. The resulting number shouldn't contain leading zeros. You can delete any number of digits between $$$0$$$ (do not delete any digits at all) and $$$n-1$$$.

For example, if you are given $$$s=$$$222373204424185217171912 then one of possible ways to make it ebne is: 222373204424185217171912 $$$\rightarrow$$$ 2237344218521717191. The sum of digits of 2237344218521717191 is equal to $$$70$$$ and is divisible by $$$2$$$, but number itself is not divisible by $$$2$$$: it means that the resulting number is ebne.

Find any resulting number that is ebne. If it's impossible to create an ebne number from the given number report about it.

Input

The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 3000$$$) — the number of digits in the original number.

The second line of each test case contains a non-negative integer number $$$s$$$, consisting of $$$n$$$ digits.

It is guaranteed that $$$s$$$ does not contain leading zeros and the sum of $$$n$$$ over all test cases does not exceed $$$3000$$$.

Output

For each test case given in the input print the answer in the following format:

- If it is impossible to create an ebne number, print "-1" (without quotes);
- Otherwise, print the resulting number after deleting some, possibly zero, but not all digits. This number should be ebne. If there are multiple answers, you can print any of them. Note that answers with leading zeros or empty strings are not accepted. It's not necessary to minimize or maximize the number of deleted digits.

Example

Input

4 4 1227 1 0 6 177013 24 222373204424185217171912

Output

1227 -1 17703 2237344218521717191

Note

In the first test case of the example, $$$1227$$$ is already an ebne number (as $$$1 + 2 + 2 + 7 = 12$$$, $$$12$$$ is divisible by $$$2$$$, while in the same time, $$$1227$$$ is not divisible by $$$2$$$) so we don't need to delete any digits. Answers such as $$$127$$$ and $$$17$$$ will also be accepted.

In the second test case of the example, it is clearly impossible to create an ebne number from the given number.

In the third test case of the example, there are many ebne numbers we can obtain by deleting, for example, $$$1$$$ digit such as $$$17703$$$, $$$77013$$$ or $$$17013$$$. Answers such as $$$1701$$$ or $$$770$$$ will not be accepted as they are not ebne numbers. Answer $$$013$$$ will not be accepted as it contains leading zeroes.

Explanation:

- $$$1 + 7 + 7 + 0 + 3 = 18$$$. As $$$18$$$ is divisible by $$$2$$$ while $$$17703$$$ is not divisible by $$$2$$$, we can see that $$$17703$$$ is an ebne number. Same with $$$77013$$$ and $$$17013$$$;
- $$$1 + 7 + 0 + 1 = 9$$$. Because $$$9$$$ is not divisible by $$$2$$$, $$$1701$$$ is not an ebne number;
- $$$7 + 7 + 0 = 14$$$. This time, $$$14$$$ is divisible by $$$2$$$ but $$$770$$$ is also divisible by $$$2$$$, therefore, $$$770$$$ is not an ebne number.

In the last test case of the example, one of many other possible answers is given. Another possible answer is: 222373204424185217171912 $$$\rightarrow$$$ 22237320442418521717191 (delete the last digit).

Codeforces (c) Copyright 2010-2020 Mike Mirzayanov

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