The main city magazine offers its readers an opportunity to publish their ads. The format of the ad should be like this:

There are space-separated non-empty words of lowercase and uppercase Latin letters.

There are hyphen characters '-' in some words, their positions set word wrapping points. Word can include more than one hyphen.

It is guaranteed that there are no adjacent spaces and no adjacent hyphens. No hyphen is adjacent to space. There are no spaces and no hyphens before the first word and after the last word.

When the word is wrapped, the part of the word before hyphen and the hyphen itself stay on current line and the next part of the word is put on the next line. You can also put line break between two words, in that case the space stays on current line. Check notes for better understanding.

The ad can occupy no more that k lines and should have minimal width. The width of the ad is the maximal length of string (letters, spaces and hyphens are counted) in it.

You should write a program that will find minimal width of the ad.

Input

The first line contains number k (1 ≤ k

The second line contains the text of the ad — non-empty space-separated words of lowercase and uppercase Latin letters and hyphens. Total length of the ad don't exceed 10⁶ characters.

Output

Output minimal width of the ad.

Examples

4
garage for sa-le

7

4
Edu-ca-tion-al Ro-unds are so fun

10


題目連結：https://vjudge.net/problem/CodeForces-803D#author=0
題意：將一個帶空格和分隔符的字串以空格或者分隔符為分割點，分割成多行字串，並且行數小於給定的k值，
而且每一行的字串長度儘可能的大，問最大的長度是多少？

思路：顯然可知：如果一個最大長度x滿足條件，x+1，x+2等等也一定滿足，即為單調的，那麼我們就可以快樂的二分了
注意下二分的區間是有限制的，總字串中可以分割成每一個小段，這些小段中最長的那個即為區間的左端點，因為比他還短一定不能存在的。
區間的右端點即為字串的總長度。
然後check函式即只需要貪心的求以mid為最長字串長度去分割要產生多少行，如果行數小於k就滿足，反而反之。
　　博主的部落格地址：https://www.cnblogs.com/qieqiemin/
我的AC程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb std::ios::sync_with_stdio(false)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int k;
string s;
vector<int> v;
int cnt;
bool check(int mid)
{
    int x;
//    int last=0;
    int m=0;
    int num=0;
    for(int i=0;i<cnt;i++)
    {
        x=v[i];
        m+=x;
        if(m>mid)
        {
            m=x;
            num++;
        }
    }
    num++;
    return num<=k;
}
int main()
{
    cin>>k;
    getchar();
    getline(cin,s);
    int len=s.length();
    int st=0;
    int pos=-1;
    for(int i=0;i<len;i++)
    {
        if(s[i]==' '||s[i]=='-')
        {
            v.push_back(i-pos);

            st=max(st,i-pos);
            pos=i;
        }
    }
    st=max(st,len-pos-1);
    v.push_back(len-pos-1);
    cnt=v.size();
//    for(int i=0;i<cnt;i)
    int l=st;
    int r=len;
    int mid;
    int ans;
    while(l<=r)
    {
        mid=(l+r)>>1;

        if(check(mid))
        {
            r=mid-1;
            ans=mid;
        }else
        {
            l=mid+1;
        }
    }
    cout<<ans<<endl;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}