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.sql.SQLSyntaxErrorException;
import java.util.*;
import java.io.*;
import java.util.stream.StreamSupport;


public class Solution {
    static long mod = (long)Math.pow(10,9)+7l;
    public static void main(String str[]) throws IOException {
//        Reader sc = new Reader();
        BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-->0){
            int n = sc.nextInt();
            sc.nextLine();
            String[] s = new String[n];
            for(int i=0;i<n;i++) {
                s[i] = sc.nextLine();
            }
            int oz = -1;
            int zo = -1;
            int ls = -1;
            int sl = -1;
            if(s[0].charAt(1)=='1') zo = 1;
            if(s[1].charAt(0)=='1') oz = 1;
            if(s[n-1].charAt(n-2)=='1') ls = 1;
            if(s[n-2].charAt(n-1)=='1') sl = 1;
            if(oz==zo && ls==sl){
                if(oz!=ls)  System.out.println(0);
                else{
                    System.out.println(2);
                    System.out.println(1+" "+2);
                    System.out.println(2+" "+1);
                }
            }
            else{
                if(oz*zo*ls*sl==-1){
                    System.out.println(1);
                    if(oz!=zo){
                        if(zo==ls){
                            System.out.println(1+" "+2);
                        }
                        else{
                            System.out.println(2+" "+1);
                        }
                    }
                    else{
                        if(ls==zo){
                            System.out.println(n+" "+(n-1));
                        }
                        else    System.out.println((n-1)+" "+n);
                    }
                }
                else{
                    System.out.println(2);
                    if(zo==ls){
                        System.out.println(1+" "+2);
                        System.out.println((n-1)+" "+n);
                    }
                    else{
                        System.out.println(1+" "+2);
                        System.out.println(n+" "+(n-1));
                    }
                }
            }
        }

        output.flush();
    }
    static int fun(ArrayList<Integer> al, int n){
        if(n==0 || n==1)    return 0;
        int LIS[] = new int[n];
        int len = 0;

        // Mark all elements of LIS as 1
        for (int i = 0; i < n; i++)
            LIS[i] = 1;

        // Find LIS of array
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (al.get(i) > al.get(j) && (i-j)<=(al.get(i)-al.get(j)))
                    LIS[i] = Math.max(LIS[i],
                            LIS[j] + 1);
            }
            len = Math.max(len, LIS[i]);
        }

        // Return min changes for array
        // to strictly increasing
        return n - len;

    }
    static long treeTraversal(Tree arr[], int parent, int x){
        long tot = 0;
        for(int i: arr[x].al){
            if(i!=parent){
                tot+=treeTraversal(arr, x, i);
            }
        }
        arr[x].child = tot;
        if(arr[x].child==0) arr[x].child = 1;
        return tot+1;
    }

    public static int primeFactors(int n, int k)
    {
        int ans = 0;
        while (n%2==0)
        {
            ans++;
            if(ans>=k)  return k;
            n /= 2;
        }
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            while (n%i == 0)
            {
                ans++;
                n /= i;
                if(ans>=k)  return k;
            }
        }

        if (n > 2)  ans++;
        return ans;
    }
    static int binaryLow(ArrayList<Integer> arr, int x, int s, int e){
        if(s>=e){
            if(arr.get(s)>=x)  return s;
            else    return s+1;
        }
        int m = (s+e)/2;
        if(arr.get(m)==x)  return m;
        if(arr.get(m)>x)   return binaryLow(arr,x,s,m);
        if(arr.get(m)<x)   return binaryLow(arr,x,m+1,e);
        return 0;
    }
    static int binaryLow(int[] arr, int x, int s, int e){
        if(s>=e){
            if(arr[s]>=x)  return s;
            else    return s+1;
        }
        int m = (s+e)/2;
        if(arr[m]==x)  return m;
        if(arr[m]>x)   return binaryLow(arr,x,s,m);
        if(arr[m]<x)   return binaryLow(arr,x,m+1,e);
        return 0;
    }
    static int binaryHigh(int[] arr, int x, int s, int e){
        if(s>=e){
            if(arr[s]<=x)  return s;
            else    return s-1;
        }
        int m = (s+e)/2;
        if(arr[m]==x)  return m;
        if(arr[m]>x)   return binaryHigh(arr,x,s,m-1);
        if(arr[m]<x)   return binaryHigh(arr,x,m+1,e);
        return 0;
    }


    static void arri(int arr[], int n, Reader sc) throws IOException{
        for(int i=0;i<n;i++){
            arr[i] = sc.nextInt();
        }
    }
    static void arrl(long arr[], int n, Reader sc) throws IOException{
        for(int i=0;i<n;i++){
            arr[i] = sc.nextLong();
        }
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    static long power(long x, long y, long p)
    {
        long res = 1; // Initialize result

        x = x % p; // Update x if it is more than or
        // equal to p

        if (x == 0)
            return 0; // In case x is divisible by p;

        while (y > 0)
        {

            // If y is odd, multiply x with result
            if ((y & 1) != 0)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }

//    static boolean dfs(Tree node, boolean[] visited, int parent, ArrayList<Tree> tt){
//        visited[node.a] = true;
//        boolean b = false;
//        for(int i: node.al){
//            if(tt.get(i).a!=parent){
//                if(visited[tt.get(i).a])    return true;
//                b|= dfs(tt.get(i), visited, node.a, tt);
//            }
//        }
//        return b;
//    }


    static class Pair{
        //        int ind;
        long a;
        long b;
//        ArrayList<Integer> ii = new ArrayList<>();
        Pair( long a, long b){
//            this.ind = i;
            this.a = a;
            this.b = b;
        }

    }
    static class SortbyI implements Comparator<Pair> {
        // Used for sorting in ascending order of
        // roll number
        public int compare(Pair a, Pair b)
        {
            if(a.b>=b.b) return 1;
            else return -1;
        }
    }
//    static class SortbyD implements Comparator<Pair> {
//        // Used for sorting in ascending order of
//        // roll number
//        public int compare(Pair a, Pair b)
//        {
//            return b.a - a.a;
//        }
//    }
    static int binarySearch(ArrayList<Pair> a, int x, int s, int e){
        if(s>=e){
            if(x<=a.get(s).b)  return s;
            else    return s+1;
        }
        int mid = (e+s)/2;
        if(a.get(mid).b<x){
            return binarySearch(a, x, mid+1, e);
        }
        else    return binarySearch(a,x,s, mid);
    }

    //    static class Edge{
//        int a;
//        int b;
//        int c;
//        int sec;
//        Edge(int a, int b, int c, int sec){
//            this.a = a;
//            this.b = b;
//            this.c = c;
//            this.sec = sec;
//        }
//
//    }
    static class Tree{
        int a;
        ArrayList<Integer> al = new ArrayList<>();
        long curr = 0;
        long freeSpace = 0;
        long child = 0;
        long max =0;
        Tree(int a){
            this.a = a;
        }

    }
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(
                    new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String readLine() throws IOException
        {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    }
                    else {
                        continue;
                    }
                }
                buf[cnt++] = (byte)c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ') {
                c = read();
            }
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0,
                    BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }


    static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;
        if (n <= 3)
            return true;
        if (n % 2 == 0 ||
                n % 3 == 0)
            return false;

        for (int i = 5;
             i * i <= n; i = i + 6)
            if (n % i == 0 ||
                    n % (i + 2) == 0)
                return false;

        return true;
    }
    static ArrayList<Integer> sieveOfEratosthenes(int n)
    {
        ArrayList<Integer> al = new ArrayList<>();
        // Create a boolean array
        // "prime[0..n]" and
        // initialize all entries
        // it as true. A value in
        // prime[i] will finally be
        // false if i is Not a
        // prime, else true.
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++)
        {
            // If prime[p] is not changed, then it is a
            // prime
            if (prime[p] == true)
            {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        // Print all prime numbers
        for (int i = 2; i <= n; i++)
        {
            if (prime[i] == true)
                al.add(i);
        }
        return al;
    }

}