time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

You are given a string s, consisting of lowercase English letters, and the integer m.

One should choose some symbols from the given string so that any contiguous subsegment of length m has at least one selected symbol. Note that here we choose positions of symbols, not the symbols themselves.

Then one uses the chosen symbols to form a new string. All symbols from the chosen position should be used, but we are allowed to rearrange them in any order.

Formally, we choose a subsequence of indices 1 ≤ i1 < i2 < … < it ≤ |s|. The selected sequence must meet the following condition: for every j such that 1 ≤ j ≤ |s| - m + 1, there must be at least one selected index that belongs to the segment [j,  j + m - 1], i.e. there should exist a k from 1 to t, such that j ≤ ik ≤ j + m - 1.

Then we take any permutation p of the selected indices and form a new string sip1sip2… sipt.

Find the lexicographically smallest string, that can be obtained using this procedure.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 100 000).

The second line contains the string s consisting of lowercase English letters. It is guaranteed that this string is non-empty and its length doesn’t exceed 100 000. It is also guaranteed that the number m doesn’t exceed the length of the string s.

Output

Print the single line containing the lexicographically smallest string, that can be obtained using the procedure described above.

Examples

input

3

cbabc

output

a

input

2

abcab

output

aab

input

3

bcabcbaccba

output

aaabb

Note

In the first sample, one can choose the subsequence {3} and form a string “a”.

In the second sample, one can choose the subsequence {1, 2, 4} (symbols on this positions are ‘a’, ‘b’ and ‘a’) and rearrange the chosen symbols to form a string “aab”.

【题解】



假如m=3;

现在给了你一段序列

从1开始。

1-3里面必然要选一个

那么选什么呢?

肯定是选字典序最小的那个。

不然的话最后选出来的一定不是最优的

比如cab

你只能选a.

然后再从xiabiao[a]+1开始再选3个

cabbb

现在下标变成3

你需要从bbb中选择一个。当然还是选择最小的那个,这个时候你没得选了

只能选b.(记录字典序最大的变成了b)

但是为了节省时间。你需要尽量往后选。

比如cabbbzab

因为如果你选了第一个b

接下来就要从b,b,z这三个里面选一个最小的。

那还是b。。所以没有意义，直接从最后一个选开始选就好。

且

b z z z b

这样的数据如果不从最后一个b开始选。你会面临3个z的抉择。这下可就会错解了。

每次选定一个字母后，就尝试更新最大字典序的字母；

最后获得了最大字典序的字母key;

从’a’..’key’-1;按顺序添加到答案字符串的末尾；

对于key要单独处理。只有在m个m个地选的过程中不得不选key这个字符的时候才要选,即加到答案字符串末尾；

因为

假如key为b

aaabb小于aaabbb

即等于key的不能全选。那样可能会出错；

而之所以小于key的要全选

是因为

aabb小于abb

即从第一个不同的字母开始比较的。

和Pascal有点不一样。pascal是先比较长度。

#include <cstdio>

#include <iostream>

#include <vector>

#include <algorithm>

#include <string>

const int MAXN = 2e5;

using namespace std;

vector <int> a[30];

string s;

bool vis[MAXN] = { 0 };

int m;

void input(int &r)

{

    r = 0;

    char t = getchar();

    while (!isdigit(t)) t = getchar();

    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();

}

int main()

{

    //freopen("F:\\rush.txt", "r", stdin);

    input(m);

    cin >> s;

    int len = s.size();

    for (int i = 0; i <= len - 1; i++)

        a[s[i] - 'a'].push_back(i);

    int i,mx = 0;

    for (i = 0; i <= len - m; i++)

    {

        for (int j = 0; j <= 25; j++)//从小到大可以肯定找的是字典序最小的。

        {

            int k = upper_bound(a[j].begin(), a[j].end(), i + m - 1) - a[j].begin();//upper_bound则尽量靠后

            k--;

            int len = a[j].size();

            if (k >= 0 && k <= len-1 && a[j][k] >= i && a[j][k] <= i + m - 1)

            {

                mx = max(mx, j);//找到之后就尝试更新最大字典序

                i = a[j][k];//等于它的位置就好第一层for会+1

                vis[a[j][k]] = true;//记录这个位置必须选

                break;//记录这个是为了方便控制输出mx字符

            }

        }

    }

    string ans = "";

    for (int i = 0; i < mx; i++)

    {

        len = a[i].size();

        for (int j = 0; j <= len - 1; j++)

            ans += (i + 'a');

    }

    len = a[mx].size();

    for (int j = 0; j <= len - 1; j++)

        if (vis[a[mx][j]])

            ans += (mx + 'a');

    cout << ans << endl;

    return 0;

}