Problem: 2^k teams participate in a playoff tournament. The tournament consists of 2^k - 1 games. They are held as follows: first of all, the teams are split into pairs: team 1 plays against team 2, team 3 plays against team 4 (exactly in this order), and so on (so, 2^{k-1} games are played in that phase). When a team loses a game, it is eliminated, and each game results in elimination of one team (there are no ties). After that, only 2^{k-1} teams remain. If only one team remains, it is declared the champion; otherwise, 2^{k-2} games are played: in the first one of them, the winner of the game "1 vs 2" plays against the winner of the game "3 vs 4", then the winner of the game "5 vs 6" plays against the winner of the game "7 vs 8", and so on. This process repeats until only one team remains.

For example, this picture describes the chronological order of games with k = 3:

<image>

Let the string s consisting of 2^k - 1 characters describe the results of the games in chronological order as follows:

  * if s_i is 0, then the team with lower index wins the i-th game; 
  * if s_i is 1, then the team with greater index wins the i-th game; 
  * if s_i is ?, then the result of the i-th game is unknown (any team could win this game). 



Let f(s) be the number of possible winners of the tournament described by the string s. A team i is a possible winner of the tournament if it is possible to replace every ? with either 1 or 0 in such a way that team i is the champion.

You are given the initial state of the string s. You have to process q queries of the following form: 

  * p c — replace s_p with character c, and print f(s) as the result of the query. 

Input

The first line contains one integer k (1 ≤ k ≤ 18).

The second line contains a string consisting of 2^k - 1 characters — the initial state of the string s. Each character is either ?, 0, or 1.

The third line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, the i-th line contains an integer p and a character c (1 ≤ p ≤ 2^k - 1; c is either ?, 0, or 1), describing the i-th query.

Output

For each query, print one integer — f(s).

Example

Input


3
0110?11
6
5 1
6 ?
7 ?
1 ?
5 ?
1 1


Output


1
2
3
3
5
4

Solution (in JAVA):

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

public class cf {
	static Reader sc = new Reader();
	static PrintWriter out = new PrintWriter(System.out);

	static int mod = (int) 1e9 + 7;

	static class Node {
		int id, val;
		long res = 0;
		Node left, right, par;

		Node(int id, int v) {
			this.id = id;
			val = v;
			left = null;
			right = null;
		}
	}

	public static void main(String[] args) throws FileNotFoundException {

		int t = 1;
		while (t-- > 0) {
			int n = sc.ni();
			char[] s = sc.next().toCharArray();
			Node[] nodes = new Node[(1 << n)];
			for (int i = 1; i < (1 << n); i++) {
				int curr = (s[i - 1] == '?' ? 2 : s[i - 1] == '1' ? 1 : 0);
				nodes[i] = new Node(i, curr);
			}
			nodes[(1 << n) - 1].par = null;
			Queue<Integer> q = new LinkedList<>();
			q.add((1 << n) - 1);
			int index = (1 << n) - 2;
			while (q.size() < (1 << (n - 1))) {
				int size = q.size();
				while (size-- > 0) {
					int curr = q.remove();
					nodes[curr].right = nodes[index];
					nodes[index].par = nodes[curr];
					q.add(index);
					index--;
					nodes[curr].left = nodes[index];
					nodes[index].par = nodes[curr];
					q.add(index);
					index--;
				}
			}

			dfs(nodes[(1 << n) - 1]);
			// System.out.println("check");
			// for (int i = 1; i < (1 << n); i++) {
			// out.print(nodes[i].res + " ");
			// }
			// out.println();

			int qq = sc.ni();
			while (qq-- > 0) {
				int i = sc.ni();
				String value = sc.next();
				int curr = value.equals("1") ? 1 : value.equals("0") ? 0 : 2;
				nodes[i].val = curr;
				update(nodes[i]);
				out.println(nodes[(1 << n) - 1].res);
			}

		}
		out.close();
	}

	private static void update(Node root) {
		if (root == null)
			return;
		int res = 0;
		if (root.left == null && root.right == null) {
			if (root.val == 1 || root.val == 0) {
				res = 1;
			} else {
				res = 2;
			}
			root.res = res;
			update(root.par);
			return;
		}

		if (root.val == 1)
			root.res = root.right.res;
		else if (root.val == 0)
			root.res = root.left.res;
		else
			root.res = root.left.res + root.right.res;

		// out.println(root.id + " " + root.res);
		update(root.par);
	}

	private static void dfs(Node root) {
		if (root.left == null && root.right == null) {
			if (root.val == 1 || root.val == 0) {
				root.res = 1;
			} else {
				root.res = 2;
			}
			return;
		}
		dfs(root.left);
		dfs(root.right);
		if (root.val == 1) {
			root.res = root.right.res;
		} else if (root.val == 0) {
			root.res = root.left.res;
		} else {
			root.res = root.left.res + root.right.res;
		}
	}

	static void shuffleSort(int[] arr) {
		// shuffle
		Random rand = new Random();
		for (int i = 0; i < arr.length; i++) {
			int j = rand.nextInt(i + 1);
			int tmp = arr[i];
			arr[i] = arr[j];
			arr[j] = tmp;
		}
		Arrays.sort(arr);
	}

	static ArrayList<Integer> sieve(int n) {
		ArrayList<Integer> primes = new ArrayList<>();
		boolean[] isPrime = new boolean[n];
		Arrays.fill(isPrime, true);
		isPrime[0] = isPrime[1] = false;
		for (int i = 2; i * i <= n; i++) {
			if (isPrime[i]) {
				for (int j = i * i; j < n; j += i) {
					isPrime[j] = false;
				}
			}
		}
		for (int i = 2; i < n; i++) {
			if (isPrime[i])
				primes.add(i);
		}
		return primes;
	}

	static class Reader {
		BufferedReader br;
		StringTokenizer st;

		Reader() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer("");
		}

		Reader(File f) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(f));
			st = new StringTokenizer("");
		}

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
				}
			return st.nextToken();
		}

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

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

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

		int[] nai(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = ni();
			return a;
		}

		long[] nal(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nl();
			return a;
		}

		char[][] nmc(int n, int m) {
			char[][] map = new char[n][m];
			for (int i = 0; i < n; i++) {
				String str = sc.next();
				for (int j = 0; j < m; j++) {
					map[i][j] = str.charAt(j);
				}
			}
			return map;
		}

		int[][] nmi(int n, int m) {
			int[][] map = new int[n][m];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					map[i][j] = ni();
				}
			}
			return map;
		}

		long[][] nml(int n, int m) {
			long[][] map = new long[n][m];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					map[i][j] = nl();
				}
			}
			return map;
		}
	}

}