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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {
	static ArrayList<Integer> adj[];

	static int k;
	static int[] a;
	static int b[];
	static int m;

	static class Pair implements Comparable<Pair> {
		long a;
		long  b;
		long c;
		public Pair(int x,int y,int z) {
			a = x;
			b = y;
			c=z;
		}
		@Override
		public int compareTo(Pair o) {
		if(o.a<this.a)
			return 1;
		return -1;
	}
}

	static ArrayList<Pair> adjlist[];
	// static char c[];
	static long mod = (long) (1e9+7);
	static int V;
	static long INF = (long) 1E16;
	static int n;
	static char c[];
	static Pair p[];
static int grid[][];
static Scanner sc = new Scanner(System.in);
static PrintWriter out = new PrintWriter(System.out);
static long div[];

public static void main(String args[]) throws Exception {
	int t=sc.nextInt();
	while(t-->0) {
	int n=sc.nextInt();
	char grid[][]=new char[n][n];
	for (int i = 0; i < grid.length; i++) {
		grid[i]=sc.next().toCharArray();
	}
	int a=grid[0][1]-'0';
	int b=grid[1][0]-'0';
	int c=grid[n-1][n-2]-'0';
	int d=grid[n-2][n-1]-'0';
	int cc=0;
	
	StringBuilder sb=new StringBuilder();
	if(a==0&&b==0) {
		if(c==0) {
		cc++;
			sb.append((n)+" "+(n-1)+"\n");
		}
		
			if(d==0) {
			cc++;
				sb.append((n-1)+" "+(n)+"\n");
			}
	}
	else if(a==1&&b==1) {
		if(c==1) {
		cc++;
			sb.append((n)+" "+(n-1)+"\n");
		}
		
			if(d==1) {
			cc++;
				sb.append((n-1)+" "+(n)+"\n");
			}
	}
	else if(a!=b) {
		if(c==d) {
		if(c==a) {
			sb.append(1+" "+2+"\n");
		cc++;
		}
		else {
			sb.append(2+" "+1+"\n");
		cc++;
		}
		}
		else {
			if(c==a) {
				sb.append(1+" "+2+"\n");
				sb.append((n-1)+" "+(n)+"\n");
				cc+=2;						
			}
			else {
				sb.append(2+" "+1+"\n");
				sb.append((n-1)+" "+(n)+"\n");
				cc+=2;						
			}
		}
		
	}
	System.out.println(cc);
	System.out.print(sb);
	}
	out.flush();
}

static int s;

	static class Query implements Comparable<Query>
	{
						// length of a block , s = sqrt(n)					
		int left, right, idx;
		int block;
		Query(int a, int b, int c)
		{
			left = a; right = b; idx = c;
			//block=left/s;
		}
		
		public int compareTo(Query q)
		{
			return right-q.right;
		}	
	}

static void dfs(int u) {
	vis[u]=true;
	for(int v:adj[u]) {
		if(!vis[v]) {
			dfs(v);
		}
	}
}

static int[] compreessarray(int c[]) {
	int b[]=new int[c.length];
	
	for (int i = 0; i < b.length; i++) {
		b[i]=c[i];
	}
	sortarray(c);
	TreeMap<Integer,Integer> map=new TreeMap<>();
	int id=0;
	for (int i = 0; i < c.length; i++) {
		if(!map.containsKey(c[i]))
			id++;
		map.put(c[i],map.getOrDefault(c[i],id));
	
	}
	int idx=0;
	for (int i = 0; i < c.length; i++) {
		joke.put(b[i],map.get(b[i]));
		b[i]=map.get(b[i]);
	}
	
	return b;
}
static TreeMap<Integer,Integer> joke=new TreeMap<>();
static long[] div4;
static long [] reversearray(long a[]) {
	for (int i = 0; i <a.length/2; i++) {
		long temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static int [] reversearray(int a[]) {
	for (int i = 0; i <=a.length/2; i++) {
		int temp=a[i];
		a[i]=a[a.length-i-1];
		a[a.length-i-1]=temp;
	}
	return a;
}
static class SegmentTree {		// 1-based DS, OOP
	
	int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
double[] array, sTree, lazy;
	
	SegmentTree(double[] in)		
	{
		array = in; N = in.length - 1;
		sTree = new double[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
		lazy = new double[N<<1];
		build(1,1,N);
	}
	
	void build(int node, int b, int e)	// O(n)
	{
		if(b == e)					
			sTree[node] = array[b];
		else						
		{
			int mid = b + e >> 1;
			build(node<<1,b,mid);
			build(node<<1|1,mid+1,e);
			sTree[node] = sTree[node<<1]+sTree[node<<1|1];
		}
	}
	
	
	void update_point(int index, double p)			// O(log n)
	{
		index += N - 1;				
		sTree[index]=p;			
		while(index>1)				
		{
			index >>= 1;
			sTree[index] = sTree[index<<1]+sTree[index<<1|1];		
		}
	}
	double query(int i, int j)
	{
		return query(1,1,N,i,j);
	}
	
	double query(int node, int b, int e, int i, int j)	// O(log n)
	{
		if(i>e || j <b)
			return 0;		
		if(b>= i && e <= j)
			return sTree[node];
		int mid = b + e >> 1;
		//propagate(node, b, mid, e);
		double q1 = query(node<<1,b,mid,i,j);
		double q2 = query(node<<1|1,mid+1,e,i,j);
		return q1+q2;	
				
	}
}
static  TreeSet<Integer> ts=new TreeSet();
static HashMap<Integer, Integer> compress(int a[]) {
		 ts = new TreeSet<>();
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : a)
			ts.add(x);
		for (int x : ts) {
			hm.put(x, hm.size() + 1);
		}
		return hm;
	}	

	static  class FenwickTree { // one-based DS

		int n;
		long[] ft;
		
		FenwickTree(int size) { n = size; ft = new long[n+1]; }
		
		long rsq(int b) //O(log n)
		{
			long sum = 0;
			while(b>0) { sum ^= ft[b];;
			 b -= b & -b;
			}		//min?
			return sum;
		}
		
		long rsq(int a, int b) { return rsq(b)^rsq(a); }	
		
		/**
		 * @param k
		 * @param l
		 */
		void point_update(int k, long l)	//O(log n), update = increment
		{
			while(k<=n) { ft[k]^=l; k += k & -k;
			}
		}
	}
	static ArrayList<Integer> euler=new ArrayList<>();
	


	
static long total;
static TreeMap<Integer,Integer> map1;

		static int zz;
//static int dp(int idx,int left,int state) {
//if(idx>=k-((zz-left)*2)||idx+1==n) {
//	return 0;
//}
//if(memo[idx][left][state]!=-1) {
//	return memo[idx][left][state];
//}
//int ans=a[idx+1]+dp(idx+1,left,0);
//if(left>0&&state==0&&idx>0) {
//	ans=Math.max(ans,a[idx-1]+dp(idx-1,left-1,1));
//}
//return memo[idx][left][state]=ans;
//}21
static HashMap<Integer,Integer> map;
static int maxa=0;
static int ff=123123;

static int[][] memo;


	static long modmod=998244353;
	static int dx[]= {1,-1,0,0};
	 static int dy[]= {0,0,1,-1};	
	static class BBOOK implements Comparable<BBOOK>{
		int t;
		int alice;
		int bob;
		public BBOOK(int x,int y,int z) {
		t=x;
		alice=y;
		bob=z;
		}
		@Override
		public int compareTo(BBOOK o) {
			return this.t-o.t;
		}
	}
	private static long lcm(long a2, long b2) {
	return (a2*b2)/gcd(a2,b2);
	}
	static class Edge implements Comparable<Edge>
	{
		int node;long cost ; long time; 
		
		Edge(int a, long b,long c) { node = a;	cost = b; time=c; }
		
		public int compareTo(Edge e){ return Long.compare(time,e.time);	}
	}
	
        static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
 
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
    static TreeSet<Integer> factors;
    static TreeSet<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
	{
		 factors = new TreeSet<Integer>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(1l*p * p <= N)
		{
			while(N % p == 0) { factors.add(p); N /= p; }
			p = primes.get(++idx);
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}

	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		long max[];
		long largest=0;

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			max=new long[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
			largest=0;
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}
		public void addmax(int i,int val) {
			max[i]=val;
			largest=Math.max(largest,max[i]);
		}
		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}
		

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];
				max[x]+=max[y];
			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;
				max[y]+=max[x];
			}
			largest=Math.max(max[x],largest);
			largest=Math.max(max[y],largest);
			
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}


	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Point  implements Comparable<Point>{
		long x, y;

		Point(long counth, long counts) {
			x = counth;
			y = counts;
		}

		@Override
		public int compareTo(Point p )
		{
			return Long.compare(p.y*1l*x, p.x*1l*y);
		}
	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/

	static boolean[] vis2;

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			int temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}


	static boolean vis[];
	static HashSet<Integer> set = new HashSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long l, long o) {
		if (o == 0) {
			return l;
		}
		return gcd(o, l % o);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			long temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
		
	}
	public static int[] sortarray(int a[]) {
		schuffle(a);
		Arrays.sort(a);
		return a;
		}
		public static long[] sortarray(long a[]) {
			schuffle(a);
			Arrays.sort(a);
			return a;
		}
		
}