Little Chris is a huge fan of linear algebra. This time he has been given a homework about the unusual square of a square matrix.

The dot product of two integer number vectors x and y of size n is the sum of the products of the corresponding components of the vectors. The unusual square of an n × n square matrix A is defined as the sum of n dot products. The i-th of them is the dot product of the i-th row vector and the i-th column vector in the matrix A.

Fortunately for Chris, he has to work only in GF(2)! This means that all operations (addition, multiplication) are calculated modulo 2. In fact, the matrix A is binary: each element of A is either 0 or 1. For example, consider the following matrix A:

<image>

The unusual square of A is equal to (1·1 + 1·0 + 1·1) + (0·1 + 1·1 + 1·0) + (1·1 + 0·1 + 0·0) = 0 + 1 + 1 = 0.

However, there is much more to the homework. Chris has to process q queries; each query can be one of the following: 

  1. given a row index i, flip all the values in the i-th row in A; 
  2. given a column index i, flip all the values in the i-th column in A; 
  3. find the unusual square of A. 



To flip a bit value w means to change it to 1 - w, i.e., 1 changes to 0 and 0 changes to 1.

Given the initial matrix A, output the answers for each query of the third type! Can you solve Chris's homework?

Input

The first line of input contains an integer n (1 ≤ n ≤ 1000), the number of rows and the number of columns in the matrix A. The next n lines describe the matrix: the i-th line contains n space-separated bits and describes the i-th row of A. The j-th number of the i-th line aij (0 ≤ aij ≤ 1) is the element on the intersection of the i-th row and the j-th column of A.

The next line of input contains an integer q (1 ≤ q ≤ 106), the number of queries. Each of the next q lines describes a single query, which can be one of the following: 

  * 1 i — flip the values of the i-th row; 
  * 2 i — flip the values of the i-th column; 
  * 3 — output the unusual square of A. 



Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.

Output

Let the number of the 3rd type queries in the input be m. Output a single string s of length m, where the i-th symbol of s is the value of the unusual square of A for the i-th query of the 3rd type as it appears in the input.

Examples

Input

3
1 1 1
0 1 1
1 0 0
12
3
2 3
3
2 2
2 2
1 3
3
3
1 2
2 1
1 1
3


Output

01001


import java.awt.List;
import java.io.*;
import java.util.*;




/*
 * kishan Thesiya
 * daiict
 * kishanthesiya25@gmail.com
 * 
 * 
 */


public class Main {
	FastScanner in;
	PrintWriter out;

	static ArrayList[] adj;
	static ArrayList[] obj;
	static int n;
	static int x;
	static int c;
	static int visited[];
	static int Ad[];
	static int Bd[];
 
	static int brr[][];
	static int a;
	static int b;
	static int parent[];
	static int size[];
	long mod = (long)1e9+7;
	static int ds[];
	
//Segment tree do not delete
	static int arr[];	
	static int d[];
	static int h;
// till now	

	class Segment{
	  private  int Lazy[];
	 
	    void ans(){
	   
	    	
	     int x = (int) (Math.ceil(Math.log(100000) / Math.log(2)));
	        int max_size = 2 * (int) Math.pow(2, x) - 1;
	        int SegTree[][] = new int[max_size][3]; // allocate memory
	        Lazy = new int[max_size];
	        
	       String s = in.next();
	       n = s.length();
	        
	        int q=in.nextInt();
	        

	        build_ST(0, 0, n - 1,SegTree,s);
	  
	        for(int i = 0;i<2*n;i++){
	        	System.out.println(SegTree[i][0] + " " + SegTree[i][1] + " " + SegTree[i][2] + " " + (i+n));
	        }
	        
	        while(q-->0)
	        {
	          int a = in.nextInt();
	          int b = in.nextInt();
	          
	          int d = getMaxST(0, n-1, a-1, b-1, 0,SegTree);
	          System.out.println(d);
	        }
	    }
	    
	    
	 void build_ST(int node, int start, int end,int SegTree[][],String s) 
	   {
	    if(start > end) return; 
	  	if(start == end) { 
	  		System.out.println(start + " " + node);
	  		if(s.charAt(start) == '('){
	  			SegTree[node][1] = 1;
	  		}else{
	  			SegTree[node][2] = 1;
	  		}
	  		
			return;
		}
		int mid=(start + end) >> 1;
		build_ST(node*2+1, start, mid, SegTree,s); // Init left child
		build_ST(node*2+2, 1+mid, end,SegTree,s); // Init right child
		
		// Init root value
		
		int t = Math.min(SegTree[2*node + 1][1], SegTree[2*node+2][2]);
		
		SegTree[node][0] = SegTree[2*node+1][0] + SegTree[2*node+2][0] + 2*t;
		SegTree[node][1] = SegTree[2*node+1][1] + SegTree[2*node+2][1] - t;
		SegTree[node][2] = SegTree[2*node+1][2] + SegTree[2*node+2][2] - t;
		
		
	}
//	    void update_ST(int node, int start, int end, int l, int r, int diff) {
//	        if(Lazy[node] != 0)
//	        {
//	            int pending = Lazy[node];
//	            
//	            SegTree[node] -= pending;
//	            
//	           if(start != end)
//	            {
//	                Lazy[2 * node + 1] += Lazy[node];
//	                Lazy[2 * node + 2] += Lazy[node];
//	            }
//	           Lazy[node] = 0;
//	        }
//	        if (end < l || start > r)
//	            return ;
//	          if(l<=start && end<=r)
//	        {
//	        	  SegTree[node] -= diff;
//	               
//	            if(start != end) // not a leaf
//	            {
//	                Lazy[2*node+1] += diff;
//	                Lazy[2*node+2] += diff;
//	            }
//	            return;
//	        }
//	        int mid=(start + end) >> 1;
//		update_ST(node*2+1, start, mid, l, r, diff); // Updating left child
//		update_ST(2+node*2, 1+mid, end, l, r, diff); // Updating right child
//	
//		SegTree[node] = Math.max(SegTree[node*2+1],SegTree[node*2+2]); 
//		
//	    }
	      int getMaxST(int start, int end, int l, int r, int node,int SegTree[][])
	    {
//	    	  if(Lazy[node] != 0)
//		        {
//		            int pending = Lazy[node];
//		            
//		            SegTree[node] -= pending;
//		            
//		           if(start != end)
//		            {
//		                Lazy[2 * node + 1] += Lazy[node];
//		                Lazy[2 * node + 2] += Lazy[node];
//		            }
//		           Lazy[node] = 0;
//		        }
	        if (l <= start && r >= end)
	            return SegTree[node][0];
	         if (end < l || start > r)
	            return 0;
	        int mid = (start + end) >> 1;
	        return getMaxST(start, mid, l, r, 2 * node + 1,SegTree)+getMaxST(mid + 1, end, l, r, 2 * node + 2,SegTree);
	    }
	      
	}
	
	static class Pair implements Comparable<Pair>
    {
		int x;int y;
		
        Pair(int l, int m) {
          
        	this.x = l;
            this.y = m;
           
        }
        
        public int compareTo(Pair p) {
            if (this.x == p.x) {
                return (int)(this.y-p.y);
            }
            return (int)(this.x - p.x);
        }
        
        public String toString(){
        	return x + " " + y;
        }
        
        public boolean equals(Object o){
        	if(o instanceof Pair){
        		Pair a = (Pair)o;
        		
        		return this.x == a.x && this.y == a.y;
        	}
        	
        	return false;
        }
        
        public int hashCode(){
        	return new Long(x).hashCode()*31 + new Long(y).hashCode();
        }
    
    }
    
	
	public static int gcd(int a, int b) {
		if (b == 0) return a;
		return gcd(b, a%b);
	}
	
	public static int find(int i){
		if(parent[i] != i){
			return find(parent[i]);
		}
		
		return i;
	}
	
	
	public static void union(int a,int b){
		int ra = find(a);
		int rb = find(b);

		if(ra!= rb){
			if(size[ra] > size[rb]){
				parent[rb] =ra;
				size[ra] += size[rb];
			}else{
				parent[ra] = rb;
				size[rb] += size[ra];
			}
		}
		
	}

	public static void apply(int p,int value){
		arr[p] += value;
		
		if(p < n){
			d[p] += value;
		}
	}
	
	public static void push(int p){
		
		for(int s = h;s>0;s--){
			int i = p >> s;
			
			if(d[i] != 0){
				apply(i<<1,d[i]);
				apply(i<<1|1,d[i]);
				d[i] = 0;
			}		
		}		
	}
	
	public static void lazybuild(int p){
		while(p > 1){
			p>>=1;
			arr[p] = Math.max(arr[p<<1], arr[p<<1|1]) + d[p];
		}
	}
	
	public static void lazyupdate(int l,int r,int value){
		l += n; r += n+1;		
		int l0 = l;int r0 = r;
		for(;l<r; l>>=1,r>>=1){
			if((l&1) == 1){
				apply(l++,value);
			}
			
			if((r&1) == 1){
				apply(--r,value);
			}
		}
		lazybuild(l0);
		lazybuild(r0-1);
			
	}
	
	public static int lazyquery(int l,int r){
		l += n;r += n+1;
		push(l);push(r-1);
		int res = (int)-2e9;
		
		for(;l<r;l>>=1,r>>=1){
			if((l&1) == 1){
				res = Math.max(arr[l++], res);
			}
			
			if((r&1) == 1){
				res = Math.max(arr[--r], res);
			}
		}
		
		return res;
	}
	
	public static void build(){
		
		for(int i = n-1;i>0;i--){
			arr[i] = Math.max(arr[i<<1], arr[i<<1|1]);
		}
		
	}
	
	public static void update(int p,int v){
		
		for(arr[p += n] = v; p>0; p>>=1 ){
			arr[p>>1] = Math.max(arr[p], arr[p^1]);
		}
		
	}
	
	public static int query(int l,int r){
		
		r++;
		int max = 0;
		for(l +=n,r +=n;l<r; l>>=1,r>>=1 ){
			
		
			
			if((l&1) == 1){
				max = Math.max(max, arr[l++]);
			}
			
			if((r&1) == 1){
				max = Math.max(max, arr[--r]);
			}
		}
		
		return max;
	}
	
	void multiply(long a[][], long b[][])
	{
	    
	    long mul[][] = new long[3][3];
	    for (int i = 0; i < 3; i++)
	    {
	        for (int j = 0; j < 3; j++)
	        {
	            mul[i][j] = 0;
	            for (int k = 0; k < 3; k++)
	                mul[i][j] = (int) ((mul[i][j] + (a[i][k]*b[k][j])%mod)%mod);
	        }
	    }
	 
	  
	    for (int i=0; i<3; i++)
	        for (int j=0; j<3; j++)
	            a[i][j] = mul[i][j];  
	}
	
	long power(long F[][], long n)
	{
	    long M[][] = {{1,1,0}, {1,0,0}, {0,1,0}};
	 
	    if (n==1)
	        return (long) ((F[0][0] + F[0][1])%mod);
	 
	    power(F, n/2);
	 
	    multiply(F, F);
	 
	    if (n%2 != 0)
	        multiply(F, M);
	 
	    return (long) ((F[0][0] + F[0][1])%mod) ;
	}
	 
	
	long findNthTerm(long n)
	{
	    long[][] F = {{1,1,0}, {1,0,0}, {0,1,0}} ;
	 
	    return power(F, n);
	}
	 
	
	
    void solve() throws NumberFormatException, IOException {
 
    	Scanner in = new Scanner (System.in);
    	
    
    	
    	
    	//System.out.println(ans);
//    	Segment s = new Segment();
//    	
//    	s.ans();
    	
//    	n = in.nextInt();
//    	arr = new int[2*n];
//    	d = new int[n];
//    	h = (int)Math.ceil((Math.log(n)/Math.log(2)));
//    	for(int i = 0;i<n;i++){
//    		arr[i+n] = in.nextInt();
//    	}
//    	
//    	build();
    	
    	 
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	
    	int n = Integer.parseInt(br.readLine());
		int c = 0;
		for(int i = 0; i < n; i++){
			String line = br.readLine();
			c += Integer.parseInt(line.split(" ")[i]);
		}
		int value = c%2;
		int q = Integer.parseInt(br.readLine());
		StringBuilder result = new StringBuilder();
		for(int i = 0; i < q; i++){
			int a = Integer.parseInt(br.readLine().split(" ")[0]);
			if(a == 1 || a == 2){
				value = (value == 1) ? 0 : 1; 
			}
			else{
				result.append(value);
			}
		}
		System.out.println(result);
    	 
//    	 StringBuilder sb = new StringBuilder();
// 
//    	 int tests = Integer.parseInt(br.readLine().trim());
//    	  
//          while (tests-- > 0) {
//        	 
//        	
//          }
//          
//          
//          System.out.println(sb.toString());
//          
           
         
}

    
    
    public static String[] reverse(String arr[]){
    	
    	
    	for(int i = 0;i<arr.length/2;i++){
    		String temp = arr[i];
    		arr[i] = arr[arr.length-i-1];
    		arr[arr.length-1-i] = temp;
    	}
    	
    	return arr;
    	
    }
	
	
    
  
    private static long ans(long n){
        if(n==1) return 2;
        n+=1;
        int cnt=-1;
        while(n%2==0){
            n/=2;
            cnt++;
        }
        if(n==1){
            return (int)Math.pow(2,cnt)-1;
        }
        return -1;
    }
    public static long sum(long x){
    	return x<10?x:x%10 + sum(x/10);
    }
	
	public static long gcd(long x, long y) {
		if (x == 0)
			return y;
		else
			return gcd( y % x,x);
	}
	public static long pow(long n,long p,long m){
        long  result = 1;
         if(p==0)
           return 1;
       if (p==1)
           return n;
       while(p!=0)
       {
           if(p%2==1)
               result *= n;
           if(result>=m)
           result%=m;
           p >>=1;
           n*=n;
           if(n>=m)
           n%=m;
       }
       return result;
       }
	
	
	
	public static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;
 
		// This is checked so that we can skip 
		// middle five numbers in below loop
		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;
	}
	

	void run() throws NumberFormatException, IOException {
		in = new FastScanner();
		out = new PrintWriter(System.out);

		solve();

		out.close();
	}

	class FastScanner {
		BufferedReader br;
		StringTokenizer st;

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

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(in.br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

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

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

		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	class Tri{
		
		Tri[] arr;
		
		public Tri(){
			arr = new Tri[27];
		}
		
	}
	
	
 

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


class Node{
	int a;
	int b;
	int c;
}

class MyComp implements Comparator<Node>{
    
	@Override
	public int compare(Node a,Node b) {
		
		// < == -1
		
		if(a.a < b.a){
			return -1;
		}else if(a.a > b.a){
			return 1;
		}else{
			return 0;
		}
		
		
	}
}

