題干：

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?≤?105).

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

Input

4 garage for sa-le

Output

7

Input

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

Output

10

Note

Here all spaces are replaced with dots.

In the first example one of possible results after all word wraps looks like this:

garage. for. sa- le

The second example:

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

題目大意：

輸入一個k和一個字符串s，最多可以把s分成k行，分割位置只能是-或者空格，求分割后最長一行長度的最小值。

解題報告：

? ?以前做過一個類似的，直接二分枚舉答案就可以了，注意fit中直接讓cnt=1，表示，當前在湊第cnt個，這樣設計cnt是為了表示最后一個點，需要多一個，所以干脆直接cnt=1好了，而且也有比較合理的解釋。

AC代碼：

#include<bits/stdc++.h>using namespace std; char s[1000005]; int ok[1000005]; int len,tot,k; bool fit(int x) {int cur = -1,cnt = 1;for(int i = 1; i<=tot; i++) {if(ok[i] - cur > x) {i--;cnt++;cur = ok[i];} if(cnt > k) return 0;}return 1; } int main() {int maxx = 0;cin>>k;getchar(); gets(s);len = strlen(s);ok[0] = -1;for(int i = 0; i<len; i++) {if(s[i] == ' ' ||s[i] == '-') {ok[++tot] = i;maxx = max(maxx,ok[tot] - ok[tot-1]);}}ok[++tot] = len-1;int l = maxx,r = len;int mid = (l + r) / 2;while(l < r) {mid = (l + r ) / 2 ;if(fit(mid)) r = mid;else l = mid + 1;} printf("%d\n",l);return 0 ;}

總結：l=maxx開始，因為這樣就避免了fit中需要先循環判斷一次是否有一個串的長度也不符合的，r從len開始，其實從哪開始都可以，剛開始寫的是len/k +1 不知道自己是怎么想的。。。肯定是錯的反正。

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