Problem: Pink Floyd are pulling a prank on Roger Waters. They know he doesn't like [walls](https://www.youtube.com/watch?v=YR5ApYxkU-U), he wants to be able to walk freely, so they are blocking him from exiting his room which can be seen as a grid.

Roger Waters has a square grid of size n× n and he wants to traverse his grid from the upper left (1,1) corner to the lower right corner (n,n). Waters can move from a square to any other square adjacent by a side, as long as he is still in the grid. Also except for the cells (1,1) and (n,n) every cell has a value 0 or 1 in it.

Before starting his traversal he will pick either a 0 or a 1 and will be able to only go to cells values in which are equal to the digit he chose. The starting and finishing cells (1,1) and (n,n) are exempt from this rule, he may go through them regardless of picked digit. Because of this the cell (1,1) takes value the letter 'S' and the cell (n,n) takes value the letter 'F'.

For example, in the first example test case, he can go from (1, 1) to (n, n) by using the zeroes on this path: (1, 1), (2, 1), (2, 2), (2, 3), (3, 3), (3, 4), (4, 4)

The rest of the band (Pink Floyd) wants Waters to not be able to do his traversal, so while he is not looking they will invert at most two cells in the grid (from 0 to 1 or vice versa). They are afraid they will not be quick enough and asked for your help in choosing the cells.  Note that you cannot invert cells (1, 1) and (n, n).

We can show that there always exists a solution for the given constraints.

Also note that Waters will pick his digit of the traversal after the band has changed his grid, so he must not be able to reach (n,n) no matter what digit he picks.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50). Description of the test cases follows.

The first line of each test case contains one integers n (3 ≤ n ≤ 200).

The following n lines of each test case contain the binary grid, square (1, 1) being colored in 'S' and square (n, n) being colored in 'F'.

The sum of values of n doesn't exceed 200.

Output

For each test case output on the first line an integer c (0 ≤ c ≤ 2) — the number of inverted cells.

In i-th of the following c lines, print the coordinates of the i-th cell you inverted. You may not invert the same cell twice.  Note that you cannot invert cells (1, 1) and (n, n).

Example

Input


3
4
S010
0001
1000
111F
3
S10
101
01F
5
S0101
00000
01111
11111
0001F


Output


1
3 4
2
1 2
2 1
0

Note

For the first test case, after inverting the cell, we get the following grid:
    
    
      
    S010  
    0001  
    1001  
    111F  
    

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.*;
import java.util.*;

// printArr() - prints the array
// intList, doubleList, longList
// nextIntArray

/* stuff you should look for
 * int overflow, array bounds
 * special cases (n=1?)
 * do smth instead of nothing and stay organized
 * WRITE STUFF DOWN
 */

public class current {
    public static void main(String[] args) {
        FastReader fr = new FastReader();
        // DO NOT FORGET TO REMOVE WHILE LOOP!!!!!!!!!!!!!!!
        // while (true) {
        int a = fr.nextInt();
        for (int i = 0; i < a; i++) {
            int n = fr.nextInt();
            char[][] arr = new char[n][n];
            for (int j = 0; j < n; j++) {
                arr[j] = fr.nextLine().toCharArray();
            }
            solve(arr);
        }
        // }
    }

    public static void solve(char[][] arr) {
        int l = arr.length;
        // b ->   0
        // a->  0
        // d ->   0
        // c -> 0
        char a = arr[1][0];
        char b = arr[0][1];
        char c = arr[l-1][l-2];
        char d = arr[l-2][l-1];
        // all same
        // 0 0 0 0
        if (c == d && a == b && c == a) {
            System.out.println(2);
            System.out.println((l-1) + " " + l);
            System.out.println(l + " " + (l-1));
        } else if (c == d && a == b) {
            // 0 0 1 1
            System.out.println(0);
        } else if (c == d || a == b) {
            // 0 1 0 0 etc
            char[] e = {a, b, c, d};
            int[] count = new int[2];
            for (char i : e) {
                if (i == '1') {
                    count[1]++;
                } else {
                    count[0]++;
                }
            }
            System.out.println(1);
            int min = Math.min(count[0], count[1]);
            char charmin = '1';
            if (min == count[0]) {
                charmin = '0';
            }
            int j = l-1;
            if (charmin == a) {
                System.out.println(1 + " " + 2);
            } else if (charmin == b) {
                System.out.println(2 + " " + 1);
            } else if (charmin == c) {
                System.out.println(j + " " + l);
            } else if (charmin == d) {
                System.out.println(l + " " + j);
            }
        } else {
            // 0 1 0 1 we still need to find a sol for
            System.out.println(2);
            // a b c d
            // 0 1 0 1
            // 1 0 1 0
            if (a == c) {
                // b & c need to cahnge
                System.out.println(1 + " " + 2);
                System.out.println(l + " " + (l-1));
            } else if (a == d) {
                // a b c d
                // 0 1 1 0
                // 1 0 0 1
                // b & d need to change
                System.out.println(1 + " " + 2);
                System.out.println((l-1) + " " + l);
            }
        }
    }

    public static Set<Integer> get(List<Set<Integer>> graph, int i) {
        try {
            return graph.get(i);
        } catch (IndexOutOfBoundsException e) {
            return new HashSet<>();
        }
    }

    public static void printArr(int[] a) {
        System.out.println(Arrays.toString(a));
    }

    public static int[] bfs(List<Set<Integer>> graph, int v) {
        int l = graph.size();
        Deque<Integer> list = new ArrayDeque<>();
        list.add(v);
        boolean[] seen = new boolean[l];
        int distance = 0;
        int node = v;
        int size = 1;
        int index = 0;
        // Iterate over every single vertice and edge
        // O(V+E) -> O(V) for trees
        while (!list.isEmpty()) {
            int cur = list.poll();
            seen[cur] = true;
            node = cur;

            index++;
            if (index > size) {
                distance++;
                size = list.size();
                index = 0;
            }

            for (int i : graph.get(cur)) {
                if (!seen[i]) {
                    seen[i] = true;
                    list.add(i);
                }
            }
        }
        return new int[]{node, distance};
    }

    static class SegTree {
        int startIndex, endIndex;
        long sum;
        SegTree lchild, rchild;

        SegTree(int[] arr) {this(0, arr.length-1, arr);}

        SegTree(int startIndex, int endIndex, int[] arr) {
            this.startIndex = startIndex;
            this.endIndex = endIndex;
            if (startIndex == endIndex) sum = arr[startIndex];
            else {
                int mid = (startIndex + endIndex) / 2;
                lchild = new SegTree(startIndex, mid, arr);
                rchild = new SegTree(mid + 1, endIndex, arr);
                sum = lchild.sum + rchild.sum;
                recalc();
            }
        }

        void recalc() {
            if (startIndex == endIndex) return; sum = lchild.sum + rchild.sum;
        }

        public void valueUpdate(int index, int value) {
            if (startIndex == endIndex) {sum = value; return;}
            if (index > lchild.endIndex) rchild.valueUpdate(index, value); else lchild.valueUpdate(index, value); recalc();
        }

        public long rangeSum(int startIndex, int endIndex) {
            if (endIndex < this.startIndex || startIndex > this.endIndex) return 0;
            if (startIndex <= this.startIndex && endIndex >= this.endIndex) return sum;
            return lchild.rangeSum(startIndex, endIndex) + rchild.rangeSum(startIndex, endIndex);
        }
    }

    public static double sum(List<Long> n) {
        double a = 0;
        for (long i : n) {
            a += i;
        }
        return a;
    }

    public static int toInt(String n) {
        return Integer.parseInt(n);
    }

    public static double toDouble(String n) {
        return Double.parseDouble(n);
    }

    public static long toLong(String n) {
        return Long.parseLong(n);
    }

    static class FastReader {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        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());
        }

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

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

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

        int[] nextIntArray() {
            return intArray(nextLine().split(" "));
        }

        long[] nextLongArray() {
            return longArray(nextLine().split(" "));
        }

        double[] nextDoubleArray() {
            return doubleArray(nextLine().split(" "));
        }

        public int getInt(int index) {
            String[] arr = nextLine().split(" ");
            return Integer.parseInt(arr[index]);
        }

        public long getLong(int index) {
            String[] arr = nextLine().split(" ");
            return Long.parseLong(arr[index]);
        }

        public double getDouble(int index) {
            String[] arr = nextLine().split(" ");
            return Double.parseDouble(arr[index]);
        }

        public List<String> stringList() {
            String[] arr = nextLine().split(" ");
            return Arrays.asList(arr);
        }

        public List<Integer> intList() {
            String[] arr = nextLine().split(" ");
            List<Integer> a = new ArrayList<>();
            for (String i : arr) {
                a.add(Integer.parseInt(i));
            }
            return a;
        }

        public List<Double> doubleList() {
            String[] arr = nextLine().split(" ");
            List<Double> a = new ArrayList<>();
            for (String i : arr) {
                a.add(Double.parseDouble(i));
            }
            return a;
        }

        public List<Long> longList() {
            String[] arr = nextLine().split(" ");
            List<Long> a = new ArrayList<>();
            for (String i : arr) {
                a.add(Long.parseLong(i));
            }
            return a;
        }
    }


    public static int[] intArray(String[] arr) {
        int l = arr.length;
        int[] a = new int[l];
        for (int i = 0; i < l; i++) {
            a[i] = Integer.parseInt(arr[i]);
        }
        return a;
    }

    public static long[] longArray(String[] arr) {
        int l = arr.length;
        long[] a = new long[l];
        for (int i = 0; i < l; i++) {
            a[i] = Long.parseLong(arr[i]);
        }
        return a;
    }

    public static double[] doubleArray(String[] arr) {
        int l = arr.length;
        double[] a = new double[l];
        for (int i = 0; i < l; i++) {
            a[i] = Double.parseDouble(arr[i]);
        }
        return a;
    }
}