You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.

The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.

Output

Print m lines, each should contain one integer — the answer to the question after the corresponding operation.

Example

Input

5
01
10
101
11111
0
3
1 2
6 5
4 4


Output

1
2
0

Note

On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.

On the second operation the string "01100" is created. Now all strings of length k = 2 are present.

On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

/**
 * @author Don Li
 */
public class HugeStrings {
    
    int M = 10;
    
    void solve() {
        int n = in.nextInt();
        char[][] s = new char[n][];
        for (int i = 0; i < n; i++) s[i] = in.nextToken().toCharArray();
        
        char[][] prefix = new char[n + 100][];
        char[][] suffix = new char[n + 100][];
        boolean[][][] has = new boolean[n + 100][M][1 << M];
        for (int i = 0; i < n; i++) {
            prefix[i] = Arrays.copyOf(s[i], Math.min(M, s[i].length));
            suffix[i] = Arrays.copyOfRange(s[i], Math.max(0, s[i].length - M), s[i].length);
            for (int len = 1; len < M; len++) {
                for (int j = 0; j <= s[i].length - len; j++) {
                    int mask = 0;
                    for (int k = 0; k < len; k++) {
                        mask = mask << 1 | (s[i][j + k] - '0');
                    }
                    has[i][len][mask] = true;
                }
            }
        }
        
        int Q = in.nextInt();
        while (Q-- > 0) {
            int x = in.nextInt() - 1, y = in.nextInt() - 1;
            for (int len = 1; len < M; len++) {
                for (int mask = 0; mask < 1 << M; mask++) {
                    has[n][len][mask] = has[x][len][mask] | has[y][len][mask];
                }
                for (int l = 1; l < len; l++) {
                    int r = len - l;
                    if (l > suffix[x].length || r > prefix[y].length) continue;
                    int mask = 0;
                    for (int i = 0; i < l; i++) {
                        mask = mask << 1 | (suffix[x][suffix[x].length - l + i] - '0');
                    }
                    for (int i = 0; i < r; i++) {
                        mask = mask << 1 | (prefix[y][i] - '0');
                    }
                    has[n][len][mask] = true;
                }
            }
            
            int ans = 0;
            L:
            for (int len = 1; len < M; len++) {
                for (int mask = 0; mask < 1 << len; mask++) {
                    if (!has[n][len][mask]) {
                        break L;
                    }
                }
                ans++;
            }
            out.println(ans);
            
            prefix[n] = Arrays.copyOf(prefix[x], Math.min(M, prefix[x].length + prefix[y].length));
            for (int i = prefix[x].length; i < prefix[n].length; i++) {
                prefix[n][i] = prefix[y][i - prefix[x].length];
            }
            suffix[n] = new char[Math.min(M, suffix[x].length + suffix[y].length)];
            for (int i = 0; i < suffix[n].length; i++) {
                if (i < suffix[y].length) suffix[n][suffix[n].length - 1 - i] = suffix[y][suffix[y].length - 1 - i];
                else suffix[n][suffix[n].length - 1 - i] = suffix[x][suffix[x].length - 1 - (i - suffix[y].length)];
            }
            n++;
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new HugeStrings().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
