The only difference between easy and hard versions is the length of the string. You can hack this problem only if you solve both problems.

Kirk has a binary string ss (a string which consists of zeroes and ones) of length nn and he is asking you to find a binary string tt of the same length which satisfies the following conditions:

• For any ll and rr (1≤l≤r≤n1≤l≤r≤n) the length of the longest non-decreasing subsequence of the substring slsl+1…srslsl+1…sr is equal to the length of the longest non-decreasing subsequence of the substring tltl+1…trtltl+1…tr;
• The number of zeroes in tt is the maximum possible.

A non-decreasing subsequence of a string pp is a sequence of indices i1,i2,…,iki1,i2,…,ik such that i1<i2<…<iki1<i2<…<ik and pi1≤pi2≤…≤pikpi1≤pi2≤…≤pik. The length of the subsequence is kk.

If there are multiple substrings which satisfy the conditions, output any.

The first line contains a binary string of length not more than 20002000.

Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.

In the first example:

• For the substrings of the length 11 the length of the longest non-decreasing subsequnce is 11;
• For l=1,r=2l=1,r=2 the longest non-decreasing subsequnce of the substring s1s2s1s2 is 1111 and the longest non-decreasing subsequnce of the substring t1t2t1t2 is 0101;
• For l=1,r=3l=1,r=3 the longest non-decreasing subsequnce of the substring s1s3s1s3 is 1111 and the longest non-decreasing subsequnce of the substring t1t3t1t3 is 0000;
• For l=2,r=3l=2,r=3 the longest non-decreasing subsequnce of the substring s2s3s2s3 is 11 and the longest non-decreasing subsequnce of the substring t2t3t2t3 is 11;

The second example is similar to the first one.