#include <bits/stdc++.h>
#define SIZE 300005
#define MOD 1000000007LL
#define INF 1 << 30
#define LLINF 1LL << 60
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define DOWN(i,b,a) for(int i=b;i>=a;i--)
#define SET(a,c) memset(a,c,sizeof a)
#define FORALL(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)
#define FOREACH(i,c) for(auto (i) : (c))
#define BIT(i,j) ((i)>>(j))&1
#define ALL(o) (o).begin(), (o).end()
#define ERASE(o) (o).erase(unique((o).begin(),(o).end()), (o).end())
#define SQ(x) ((x)*(x))
using namespace std;
typedef long long ll;
typedef pair<ll,ll> Pll;
typedef pair<int, int> Pii;
typedef pair<double, double> Pdd;
typedef complex<double> dcomplex;
template<typename T> inline void priv(vector<T>a){REP(i,a.size()){cerr<<a[i]<<((i==a.size()-1)?"\n":" ");}}
ll gcd(ll a,ll b){int c=max(a,b);int d=min(a,b);return c==0||d==0?c:gcd(c%d,d);}
ll lcm(ll a,ll b){return a==0||b==0?0:a*b/gcd(a,b);}
ll fact(ll a){ll b=1;FOR(i,1,a)b*=i;return b;}

int n;
int a[10][10];

bool big(int i, int j)
{
  FOR(k,-2,2) FOR(l,-(2-abs(k)),(2-abs(k))) if(a[i+k][j+l]==0) return false;
  return true;
}

bool medium(int i, int j)
{
  FOR(k,-1,1) FOR(l,-1,1) if(a[i+k][j+l]==0) return false;
  return true;
}

bool small(int i, int j)
{
  FOR(k,-1,1) FOR(l,-(1-abs(k)),1-abs(k)) if(a[i+k][j+l]==0) return false;
  return true;
}

bool dfs(int n)
{
  if(n==0)
  {
    REP(i,10) REP(j,10) if(a[i][j] > 0) return false;
    return true;
  }
  FOR(i,2,7) FOR(j,2,7) if(big(i,j))
  {
    FOR(k,-2,2) FOR(l,-(2-abs(k)),(2-abs(k))) a[i+k][j+l]--;
    if(dfs(n-1)) {printf("%d %d %d\n",j,i,3); return true;}
    FOR(k,-2,2) FOR(l,-(2-abs(k)),(2-abs(k))) a[i+k][j+l]++;
  }
  FOR(i,1,8) FOR(j,1,8) if(medium(i,j))
  {
    FOR(k,-1,1) FOR(l,-1,1) a[i+k][j+l]--;
    if(dfs(n-1)) {printf("%d %d %d\n",j,i,2); return true;}
    FOR(k,-1,1) FOR(l,-1,1) a[i+k][j+l]++;
  }
  FOR(i,1,8) FOR(j,1,8) if(small(i,j))
  {
    FOR(k,-1,1) FOR(l,-(1-abs(k)),(1-abs(k))) a[i+k][j+l]--;
    if(dfs(n-1)) {printf("%d %d %d\n",j,i,1); return true;}
    FOR(k,-1,1) FOR(l,-(1-abs(k)),(1-abs(k))) a[i+k][j+l]++;
  }
  return false;
}

int main()
{
  cin >> n;
  REP(i,10) REP(j,10) cin >> a[i][j];
  dfs(n);
  return 0;
}