【April Fools Day Contest 2016E】【脑洞 define分行用法】Out of Controls 不用条件循环实现最长最短路

E. Out of Controls

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

You are given a complete undirected graph. For each pair of vertices you are given the length of the edge that connects them. Find the shortest paths between each pair of vertices in the graph and return the length of the longest of them.

Input

The first line of the input contains a single integer N (3 ≤ N ≤ 10).

The following N lines each contain N space-separated integers. jth integer in ith line a_ij is the length of the edge that connects vertices iand j. a_ij = a_ji, a_ii = 0, 1 ≤ a_ij ≤ 100 for i ≠ j.

Output

Output the maximum length of the shortest path between any pair of vertices in the graph.

Examples

input

3
0 1 1
1 0 4
1 4 0

output

2

input

4
0 1 2 3
1 0 4 5
2 4 0 6
3 5 6 0

output

5

Note

You're running short of keywords, so you can't use some of them:

define
do
for
foreach
while
repeat
until
if
then
else
elif
elsif
elseif
case
switch

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
int n;
int QAQ;
int a[12][12];
bool read(int i, int j)
{
	scanf("%d", &a[i][j]);
	j < n&&read(i, j + 1);
	j == n&&i < n&&read(i + 1, 1);
	return 1;
}
bool floyd(int k, int i, int j)
{
	k < n&&i == n&&j == n&&floyd(k + 1, 1, 1);
	i < n&&j == n&&floyd(k, i + 1, 1);
	j < n&&floyd(k, i, j + 1);
	a[i][j] = min(a[i][j], a[i][k] + a[k][j]);
	return 1;
}
int ans;
bool update(int i, int j)
{
	ans = max(ans, a[i][j]);
	j < n&&update(i, j + 1);
	j == n&&i < n&&update(i + 1, 1);
	return 1;
}
int main()
{
	scanf("%d", &n);
	read(1, 1);
	floyd(1, 1, 1);
	ans = 0;
	update(1, 1);
	printf("%d\n", ans);
	return 0;
}
/*
在define语句中，\可以断2行，可以更方便地AC
*/