Problem:
Carpenters' Language

International Carpenters Professionals Company (ICPC) is a top construction company with a lot of expert carpenters.
What makes ICPC a top company is their original language.

The syntax of the language is simply given in CFG as follows:

S -> SS | (S) | )S( | ε

In other words, a right parenthesis can be closed by a left parenthesis and a left parenthesis can be closed by a right parenthesis in this language.

Alex, a grad student mastering linguistics, decided to study ICPC's language.
As a first step of the study, he needs to judge whether a text is well-formed in the language or not.
Then, he asked you, a great programmer, to write a program for the judgement.

Alex's request is as follows: You have an empty string S in the beginning, and construct longer string by inserting a sequence of '(' or ')' into the string.
You will receive q queries, each of which consists of three elements (p, c, n), where p is
the position to insert, n is the number of characters to insert and c is either '(' or ')', the character to insert.
For each query, your program should insert c repeated by n times into the p-th position of S from the beginning.
Also it should output, after performing each insert operation, "Yes" if S is in the language and "No" if S is not in the language.

Please help Alex to support his study, otherwise he will fail to graduate the college.

Input

The first line contains one integer q (1 \leq q \leq 10^5) indicating the number of queries, follows
q lines of three elements, p_i, c_i, n_i, separated by a single space (1 \leq i \leq q, c_i = '(' or ')',
0 \leq p_i \leq length of S before i-th query, 1 \leq n \leq 2^{20}).
It is guaranteed that all the queries in the input are valid.

Output

For each query, output "Yes" if S is in the language and "No" if S is not in the language.

Sample Input 1

3
0 ( 10
10 ) 5
10 ) 5

Output for the Sample Input 1

No
No
Yes

Sample Input 2

3
0 ) 10
10 ( 5
10 ( 5

Output for the Sample Input 2

No
No
Yes

Sample Input 3

3
0 ( 10
10 ) 20
0 ( 10

Output for the Sample Input 3

No
No
Yes

Solution in JAVA (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;
import java.awt.geom.*;
import java.math.*;

public class Main {

	static final Scanner in = new Scanner(System.in);
	static final PrintWriter out = new PrintWriter(System.out,false);
	static boolean debug = false;

	static boolean isLang(String s) {
		int n = s.length();
		if (n == 2) {
			if (s.charAt(0) == '(' && s.charAt(n-1) == ')') return true;
			if (s.charAt(0) == ')' && s.charAt(n-1) == '(') return true;
		}
		if (n == 1) return false;
		if (n == 0) return true;
		if ((s.charAt(0) == '(' && s.charAt(n-1) == ')') || (s.charAt(0) == ')' && s.charAt(n-1) == '(')) {
			if (isLang(s.substring(1, n-1))) return true;
		}
		for (int i=1; i<n-1; i++) {
			if (isLang(s.substring(0, i)) && isLang(s.substring(i, n)))
				return true;
		}
		return false;
	}

	static void solve() {
		int q = in.nextInt();
		StringBuilder s = new StringBuilder();
		while (q-- > 0) {
			int pos = in.nextInt();
			char c = in.next().charAt(0);
			int t = in.nextInt();
			while (t-- > 0) {
				s.insert(pos, c);
			}
			out.println(isLang(s.toString()) ? "Yes" : "No");
		}
	}

	public static void main(String[] args) {
		debug = args.length > 0;
		long start = System.nanoTime();

		solve();
		out.flush();

		long end = System.nanoTime();
		dump((end - start) / 1000000 + " ms");
		in.close();
		out.close();
	}

	static void dump(Object... o) { if (debug) System.err.println(Arrays.deepToString(o)); }
}