A tournament is a directed graph without self-loops in which every pair of vertexes is connected by exactly one directed edge. That is, for any two vertexes $*u*$ and $*v*$ ($*u* ≠ *v*$) exists either an edge going from $*u*$ to $*v*$, or an edge from $*v*$ to $*u*$.

You are given a tournament consisting of $*n*$ vertexes. Your task is to find there a cycle of length three.

Input

The first line contains an integer $*n*$ ($1 ≤ *n* ≤ 5000$). Next $*n*$ lines contain the adjacency matrix $*A*$ of the graph (without spaces). $*A*_{*i*, *j*} = 1$ if the graph has an edge going from vertex $*i*$ to vertex $*j*$, otherwise $*A*_{*i*, *j*} = 0$. $*A*_{*i*, *j*}$ stands for the $*j*$-th character in the $*i*$-th line.

It is guaranteed that the given graph is a tournament, that is, $*A*_{*i*, *i*} = 0, *A*_{*i*, *j*} ≠ *A*_{*j*, *i*}$ $(1 ≤ *i*, *j* ≤ *n*, *i* ≠ *j*)$.

Output

Print three distinct vertexes of the graph $*a*_{1}$, $*a*_{2}$, $*a*_{3}$ ($1 ≤ *a*_{*i*} ≤ *n*$), such that $*A*_{*a*_{1}, *a*_{2}} = *A*_{*a*_{2}, *a*_{3}} = *A*_{*a*_{3}, *a*_{1}} = 1$, or "-1", if a cycle whose length equals three does not exist.

If there are several solutions, print any of them.