#include <bits/stdc++.h>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
const double EPS = 1e-9;
int v[10];
int a[10][5];
vector<vector<int>> permutations;
double dist(const pair<pii, int> &a, const pair<pii, int> &b) {
  return sqrt(1.0 * (a.first.first - b.first.first) *
                  (a.first.first - b.first.first) +
              1.0 * (a.first.second - b.first.second) *
                  (a.first.second - b.first.second) +
              1.0 * (a.second - b.second) * (a.second - b.second));
}
void check() {
  vector<pair<pii, int>> b;
  for (int i = 0; i < 8; ++i) {
    const auto &p = permutations[v[i]];
    b.push_back({{a[i][p[0]], a[i][p[1]]}, a[i][p[2]]});
  }
  for (int i = 0; i < 8; ++i)
    for (int j = i + 1; j < 8; ++j)
      if (b[i] == b[j]) return;
  auto c1 = *min_element(b.begin(), b.end());
  auto c2 = *max_element(b.begin(), b.end());
  double aux = dist(c1, c2) / sqrt(3.0);
  if (abs(aux - round(aux)) > EPS) return;
  int d = round(aux);
  vector<int> v1, v2;
  for (int i = 0; i < 8; ++i) {
    if (abs(dist(c1, b[i]) - d) < EPS) v1.push_back(i);
    if (abs(dist(c2, b[i]) - d) < EPS) v2.push_back(i);
  }
  if (v1.size() != 3 || v2.size() != 3) return;
  cout << "YES\n";
  for (int i = 0; i < 8; ++i)
    cout << b[i].first.first << ' ' << b[i].first.second << ' ' << b[i].second
         << '\n';
  exit(0);
}
void bkt(int k) {
  if (k == 8) {
    check();
    return;
  }
  for (int i = 0; i < permutations.size(); ++i) {
    v[k] = i;
    bkt(k + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  vector<int> p = {0, 1, 2};
  do {
    permutations.push_back(p);
  } while (next_permutation(p.begin(), p.end()));
  for (int i = 0; i < 8; ++i)
    for (int j = 0; j < 3; ++j) cin >> a[i][j];
  bkt(1);
  cout << "NO\n";
  return 0;
}
