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:

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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.text.*;

public class A {

	public A () {
		int N = sc.nextInt();
		Integer [][] A = sc.nextInts(N);

		int res = 0;
		for (int i : rep(N))
			res += A[i][i];
		res %= 2;

		StringBuilder b = new StringBuilder();
		for (@SuppressWarnings("unused") int q : rep(sc.nextInt()))
			if (sc.nextInt() == 3)
				b.append(res);
			else {
				sc.nextInt();
				++res;
				res %= 2;
			}
		exit(b);
	}

	private static long ceil (long p, long q) {
		if (q < 0) return ceil(-p, -q);
		else if (p < 0) return -floor(-p, q);
		else return (p+q-1) / q;
	}
	private static long floor(long p, long q) {
		if (q < 0) return floor(-p, -q);
		else if (p < 0) return -ceil(-p, q);
		else return p / q;
	}
	private static int [] rep(int N) { return rep(0, N); }
	private static int [] rep(int S, int T) { if (T <= S) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	////////////////////////////////////////////////////////////////////////////////////
	private final static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static void print (Object o, Object ...a) { IOUtils.print(o, a); }
	private static void exit (Object o, Object ...a) { print(o, a); exit(); }
	private static void exit() { IOUtils.exit(); }
	private static class IOUtils {
		public static class MyScanner {
			public String next() { newLine(); return line[index++]; }
			public int nextInt() { return Integer.parseInt(next()); }
			public String nextLine() { line = null; return readLine(); }
			public String [] nextStrings() { return split(nextLine()); }
			public Integer [] nextInts() {
				String [] L = nextStrings();
				Integer [] res = new Integer [L.length];
				for (int i = 0; i < L.length; ++i)
					res[i] = Integer.parseInt(L[i]);
				return res;
			}
			public Integer [][] nextInts (int N) {
				Integer [][] res = new Integer [N][];
				for (int i = 0; i < N; ++i)
					res[i] = nextInts();
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol() { return index == line.length; }
			private String readLine() {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine() {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split(String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build(Object o, Object... a) { return buildDelim(" ", o, a); }
		private static String buildDelim(String delim, Object o, Object... a) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : a)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static void start() { if (t == 0) t = millis(); }
		private static void append(StringBuilder b, Object o, String delim) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					append(b, java.lang.reflect.Array.get(o, i), delim);
			} else if (o instanceof Iterable<?>)
				for (Object p : (Iterable<?>) o)
					append(b, p, delim);
			else {
				if (o instanceof Double)
					o = new DecimalFormat("#.############").format(o);
				b.append(delim).append(o);
			}
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void print(Object o, Object ...a) { pw.println(build(o, a)); }
		private static void err(Object o, Object ...a) { System.err.println(build(o, a)); }
		private static void exit() {
			IOUtils.pw.close();
			System.out.flush();
			err("------------------");
			err(IOUtils.time());
			System.exit(0);
		}
		private static long t;
		private static long millis() { return System.currentTimeMillis(); }
		private static String time() { return "Time: " + (millis() - t) / 1000.0; }
	}
	public static void main (String[] args) { new A(); exit(); }
}
