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.awt.Point;
import java.awt.geom.Point2D;
import java.util.Comparator;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Easy2 {
    static char[] alphabet = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y','Z'};
    static long mod = (long) (1e9 + 7);
    static int[] dx = {1,-1,0,0};
    static int[] dy = {0,0,1,-1};
    static int[] ddx = {1,-1,0,0,1,-1,1,-1};
    static int[] ddy = {0,0,1,-1,1,-1,-1,1};
    static double eps = 1e-6;

    static int[] dist;
    static boolean[] trav;
    static Node[] nodes;
    public static void main(String[] args) throws Exception {
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt();
        HashMap<PointL, Long> map = new HashMap<>();
        HashMap<PointL, Integer> ind = new HashMap<>();
        int[] t1 = {0,0,1};
        int[] t2 = {1,2,2};
        int[] t3 = {2,1,0};
        //int[] t1 = {0,0,1,1,2,2};
        //int[] t2 = {1,2,2,0,0,1};
        //int[] t3 = {2,1,0,2,1,0};
        long best = 0;
        long best1 = -1;
        long best2 = -1;
        long[] all = new long[3];
        for (int i = 0; i < n; i++) {
            all[0] = sc.nextLong();
            all[1] = sc.nextLong();
            all[2] = sc.nextLong();
            long min = Math.min(all[2], Math.min(all[0], all[1]));
            if (min > best) {
                best = min;
                best1 = i;
                best2 = -1;
            }

            for (int j = 0; j < 3; j++) {
                min = Math.min(all[t1[j]], all[t2[j]]);
                if (min > best) {
                    PointL curr = new PointL(all[t1[j]], all[t2[j]]);
                    if (map.containsKey(curr)) {
                        long last = map.get(curr);
                        min = Math.min(min, last + all[t3[j]]);
                        if (min > best) {
                            best = min;
                            best1 = i;
                            best2 = ind.get(curr);
                        }
                        if (map.get(curr) < all[t3[j]]) {
                            map.put(curr, all[t3[j]]);
                            ind.put(curr, i);
                        }
                    } else {
                        map.put(curr, all[t3[j]]);
                        ind.put(curr, i);
                    }
                }
            }
        }
        //System.out.println(best);
        out.println(best2 == -1 ? "1" : "2");
        out.print(best1 + 1);
        if (best2 != -1) out.println(" " + (best2 + 1));
        else out.println();
        out.close();
    }

    public static class PointL {
        long a;
        long b;

        public PointL(long a, long b) {
            this.a = a;
            this.b = b;
        }

        public boolean equals(Object o) {
            if (!(o instanceof PointL)) return false;
            PointL o2 = (PointL) o;
            return (a == o2.a && b == o2.b) || (a == o2.b && b == o2.a);
        }

        public int hashCode() {
            return Long.hashCode(a) ^ Long.hashCode(b);
        }
    }

    public static class Node {
        HashMap<Integer, Long> edges;
        public Node() {
            edges = new HashMap<>();
        }
    }
    public static class STree {
        long[] tree;
        long[] base;
        public STree(long[] base) {
            this.base = base;
            tree = new long[2 * base.length];
            for (int i = base.length; i < tree.length; i++) {
                tree[i] = base[i - base.length];
            }
            for (int i = base.length - 1; i > 0; i--) {
                tree[i] = tree[2*i] + tree[2*i ^ 1];
            }
        }

        public long get(int index) {
            return tree[index + base.length];
        }

        public void modify(int index, long val) {
            index += base.length;
            tree[index] = val;
            while (index > 1) {
                tree[index/2] = tree[index] + tree[index ^ 1];
                index/=2;
            }
        }

        public long query(int start, int end) {
            long ans = 0;
            start += base.length;
            end += base.length + 1;
            while (start < end) {
                if ((start & 1) == 1) {
                    ans += tree[start++];
                }
                if ((end & 1) == 1) {
                    ans += tree[--end];
                }
                start>>=1;
            end>>=1;
            }
            return ans;
        }

        public void print() {
            out.println();
            for (int i = base.length; i < tree.length; i++) {
                out.print(tree[i] + " ");
            }
            out.println();
        }
    }

    public static PrintWriter out;
    /*
    int n      = sc.nextInt();        // read input as integer
    long k     = sc.nextLong();       // read input as long
    double d   = sc.nextDouble();     // read input as double
    String str = sc.next();           // read input as String
    String s   = sc.nextLine();       // read whole line as String
    out.println(result);              // print via PrintWriter
     */
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
}