Problem: Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya's idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere. 

Zahar has n stones which are rectangular parallelepipeds. The edges sizes of the i-th of them are ai, bi and ci. He can take no more than two stones and present them to Kostya. 

If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way. 

Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.

Input

The first line contains the integer n (1 ≤ n ≤ 105).

n lines follow, in the i-th of which there are three integers ai, bi and ci (1 ≤ ai, bi, ci ≤ 109) — the lengths of edges of the i-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.

Output

In the first line print k (1 ≤ k ≤ 2) the number of stones which Zahar has chosen. In the second line print k distinct integers from 1 to n — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to n in the order as they are given in the input data.

You can print the stones in arbitrary order. If there are several answers print any of them. 

Examples

Input

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


Output

1
1


Input

7
10 7 8
5 10 3
4 2 6
5 5 5
10 2 8
4 2 1
7 7 7


Output

2
1 5

Note

In the first example we can connect the pairs of stones:

  * 2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
  * 2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5 respectively. 
  * 2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5
  * 4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1
  * 5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5



Or take only one stone:

  * 1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5
  * 2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
  * 3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5
  * 4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5
  * 5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1
  * 6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5



It is most profitable to take only the first stone. 

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.util.*;
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Supplier;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main {
  void run() {
    int n = ni();
    int[][] rec = new int[n + 1][4];
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j < 3; ++j) {
        rec[i][j] = ni();
      }
      rec[i][3] = i;
      Arrays.sort(rec[i], 0, 3);
    }
    int[][] case1 = Stream.of(rec).map(a -> IntStream.of(a).toArray()).toArray(int[][]::new);
    Comparator<int[]> comp1 = (int[] a, int[] b) -> {
      for (int i = 0; i < 3; ++i) {
        if (i == 0) {
          continue;
        }
        if (a[i] != b[i]) {
          return a[i] - b[i];
        }
      }
      return a[0] - b[0];
    };
    Arrays.sort(case1, comp1);
    int[][] case2 = Stream.of(rec).map(a -> IntStream.of(a).toArray()).toArray(int[][]::new);
    Comparator<int[]> comp2 = (int[] a, int[] b) -> {
      for (int i = 0; i < 3; ++i) {
        if (i == 1) {
          continue;
        }
        if (a[i] != b[i]) {
          return a[i] - b[i];
        }
      }
      return a[1] - b[1];
    };
    Arrays.sort(case2, comp2);
    int[][] case3 = Stream.of(rec).map(a -> IntStream.of(a).toArray()).toArray(int[][]::new);
    Comparator<int[]> comp3 = (int[] a, int[] b) -> {
      for (int i = 0; i < 3; ++i) {
        if (i == 2) {
          continue;
        }
        if (a[i] != b[i]) {
          return a[i] - b[i];
        }
      }
      return a[2] - b[2];
    };
    Arrays.sort(case3, comp3);
    int max = 0;
    int max_idx = 0;
    for (int i = 1; i <= n; ++i) {
      int min = rec[i][0];
      if (max < min) {
        max = min;
        max_idx = i;
      }
    }

    boolean flag = false;
    int[] ans = new int[2];
    for (int i = 1; i <= n; ++i) {
      int min = rec[i][0];
      int next_min = rec[i][1];
      int[] arr = new int[3];
      Arrays.fill(arr, Integer.MAX_VALUE);
      for (int j = 1; j < 3; ++j) {
        arr[j] = rec[i][j];
      }
      int idx1 = ~Arrays.binarySearch(case1, arr, comp1);
      --idx1;
      if (case1[idx1][3] == i) {
        --idx1;
      }
//      debug(i, arr, idx1, case1);
      int v1 = Math.min(case1[idx1][0] + min, next_min);
      if (case1[idx1][1] == arr[1] &&
          case1[idx1][2] == arr[2] &&
          v1 > max) {
        max = v1;
        flag = true;
        ans[0] = i;
        ans[1] = case1[idx1][3];
      }
      arr[0] = arr[1];
      arr[1] = Integer.MAX_VALUE;
      int idx2 = ~Arrays.binarySearch(case2, arr, comp2);
      --idx2;
      if (case1[idx2][3] == i) {
        --idx2;
      }
      int v2 = Math.min(case1[idx2][1] + min, next_min);
      if (case1[idx2][0] == arr[0] &&
          case1[idx2][2] == arr[2] &&
          case1[idx2][3] != i && v2 > max) {
        max = v2;
        flag = true;
        ans[0] = i;
        ans[1] = case1[idx2][3];
      }
      arr[1] = arr[2];
      arr[2] = Integer.MAX_VALUE;
      int idx3 = ~Arrays.binarySearch(case3, arr, comp3);
      --idx3;
      if (case1[idx3][3] == i) {
        --idx3;
      }
      int v3 = Math.min(case1[idx3][2] + min, next_min);
      if (case1[idx2][0] == arr[0] &&
          case1[idx2][1] == arr[1] &&
          case1[idx3][3] != i && v3 > max) {
        max = v3;
        flag = true;
        ans[0] = i;
        ans[1] = case1[idx3][3];
      }
    }

    Arrays.sort(ans);
    if (flag) {
      System.out.println(2);
      System.out.println(ans[0] + " " + ans[1]);
    } else {
      System.out.println(1);
      System.out.println(max_idx);
    }
  }

  Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    new Main().run();
  }

  int ni() {
    return Integer.parseInt(sc.next());
  }

  void debug(Object... os) {
    System.err.println(Arrays.deepToString(os));
  }

  class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    /**
     * 1-indexed なBinary Indexed Treeを構築する
     *
     * @param n   容量
     * @param bif 適用させる関数
     * @param sup 初期値
     */
    BIT(int n, BiFunction<T, T, T> bif, Supplier<T> sup) {
      this.n = n;
      bit = new ArrayList<>(n + 1);
      for (int i = 0; i < n + 1; ++i) {
        bit.add(sup.get());
      }
      this.bif = bif;
    }

    /**
     * iの位置の値をvで更新する
     *
     * @param i index
     * @param v 新しい値
     */
    void update(int i, T v) {
      for (int x = i; x <= n; x += x & -x) {
        bit.set(x, bif.apply(bit.get(x), v));
      }
    }

    /**
     * クエリー
     *
     * @param defaultValue 初期値
     * @param i            index
     * @return [1, i]までfを適用した結果
     */
    T reduce(T defaultValue, int i) {
      T ret = defaultValue;
      for (int x = i; x > 0; x -= x & -x) {
        ret = bif.apply(ret, bit.get(x));
      }
      return ret;
    }
  }

  long MOD = 1_000_000_007;

  /**
   * 繰り返し2乗法を用いたべき乗の実装
   *
   * @return a^r (mod 1,000,000,007)
   */
  long pow(long a, long r) {
    long sum = 1;
    while (r > 0) {
      if ((r & 1) == 1) {
        sum *= a;
        sum %= MOD;
      }
      a *= a;
      a %= MOD;
      r >>= 1;
    }
    return sum;
  }

  /**
   * 組み合わせ
   * O(n)
   *
   * @return {}_nC_r
   */
  long C(int n, int r) {
    long sum = 1;
    for (int i = n; 0 < i; --i) {
      sum *= i;
      sum %= MOD;
    }
    long s = 1;
    for (int i = r; 0 < i; --i) {
      s *= i;
      s %= MOD;
    }
    sum *= pow(s, MOD - 2);
    sum %= MOD;

    long t = 1;
    for (int i = n - r; 0 < i; --i) {
      t *= i;
      t %= MOD;
    }
    sum *= pow(t, MOD - 2);
    sum %= MOD;

    return sum;
  }

  double GOLDEN_RATIO = (1.0 + Math.sqrt(5)) / 2.0;

  /**
   * 黄金分割探索
   *
   * @param left  下限
   * @param right 上限
   * @param f     探索する関数
   * @param comp  上に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue)
   *              下に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue).reversed()
   * @return 極値の座標x
   */
  double goldenSectionSearch(double left, double right, Function<Double, Double> f, Comparator<Double> comp) {
    double c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
    double c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
    double d1 = f.apply(c1);
    double d2 = f.apply(c2);
    while (right - left > 1e-9) {
      if (comp.compare(d1, d2) > 0) {
        right = c2;
        c2 = c1;
        d2 = d1;
        c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
        d1 = f.apply(c1);
      } else {
        left = c1;
        c1 = c2;
        d1 = d2;
        c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
        d2 = f.apply(c2);
      }
    }
    return right;
  }

  /**
   * [a,b]をm:nに内分する点を返す
   */
  double divideInternally(double a, double b, double m, double n) {
    return (n * a + m * b) / (m + n);
  }

  long gcd(long a, long b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }

  int gcd(int a, int b) {
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }

  long lcd(long a, long b) {
    return (a / gcd(a, b)) * b;
  }

  int lcd(int a, int b) {
    return (a / gcd(a, b)) * b;
  }

  /**
   * http://qiita.com/p_shiki37/items/65c18f88f4d24b2c528b
   */
  static class FastScanner {
    private final InputStream in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    public FastScanner(InputStream in) {
      this.in = in;
    }

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte()) return buffer[ptr++];
      else return -1;
    }

    private static boolean isPrintableChar(int c) {
      return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
      while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
    }

    public boolean hasNext() {
      skipUnprintable();
      return hasNextByte();
    }

    public String next() {
      if (!hasNext()) throw new NoSuchElementException();
      StringBuilder sb = new StringBuilder();
      int b = readByte();
      while (isPrintableChar(b)) {
        sb.appendCodePoint(b);
        b = readByte();
      }
      return sb.toString();
    }

    public long nextLong() {
      if (!hasNext()) throw new NoSuchElementException();
      long n = 0;
      boolean minus = false;
      int b = readByte();
      if (b == '-') {
        minus = true;
        b = readByte();
      }
      if (b < '0' || '9' < b) {
        throw new NumberFormatException();
      }
      while (true) {
        if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
        } else if (b == -1 || !isPrintableChar(b)) {
          return minus ? -n : n;
        } else {
          throw new NumberFormatException();
        }
        b = readByte();
      }
    }
  }
}