D1. Kirk and a Binary String (easy version)

题目链接：http://codeforces.com/contest/1204/problem/D1

D1. Kirk and a Binary String (easy version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output

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

Examples

input

Copy

110

output

Copy

010

input

Copy

010

output

Copy

010

input

Copy

0001111

output

Copy

0000000

input

Copy

0111001100111011101000

output

Copy

0011001100001011101000

Note

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.

题目大意：

给出一个 串 S ,求一个串 T  要求，等长所有区间的 LIS（最长上升子序列） 相等，0 的个数尽可能多

题解：

从后向前，保证后面的解都是合法的情况下
如果当前位置的数字是 0
那么，他一定是后面以他为起点的区间的 LIS LISLIS 的一部分，这就要求 T TT 的对应位置必须为 0, 否则 LIS LISLIS 长度必然减少

如果当前位置的数字为 1
考虑，以他为起点的所有区间
对于那些 LIS LISLIS 包含它的区间，就是说 LIS LISLIS 的首项为 1 的区间，他变为0，对这些区间的 LIS LISLIS 没有影响（他们的 LIS LISLIS 长度为 1 的个数）
对于那些 LIS LISLIS 不包含他的区间，就是说 LIS LISLIS 的首项为 0 的区间，他变为0，对这些区间的 LIS LISLIS 会改变

换句话说，若想将 1 变为 0 ，必须保证后面所有的区间的 LIS LISLIS 长度必须和 1 的个数相等！！！

所以，从后向前统计 0 和 1 的数量，当 1 的个数大于等于 0 的个数时，才可以修改

代码：判断后面是否所有区间的1的个数大于0的个数竟然可以O（1）难以想象，一直以为要一遍for才行的

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    string s;cin>>s;
    int len=s.size();
    int sum=0;
    for(int i=len-1;i>=0;i--)
    {
        if(s[i]=='0') sum++;
        else if(sum) sum--;
        else s[i]='0';
    }
    cout<<s<<endl;
    return 0;
}