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.

Examples

Input

3
0 ( 10
10 ) 5
10 ) 5


Output

No
No
Yes


Input

3
0 ) 10
10 ( 5
10 ( 5


Output

No
No
Yes


Input

3
0 ( 10
10 ) 20
0 ( 10


Output

No
No
Yes
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <cfloat>
#include <ctime>
#include <cassert>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <numeric>
#include <list>


using namespace std;

#ifdef _MSC_VER
#define __typeof__ decltype
template <class T> int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }
#endif

#define foreach(it, c) for (__typeof__((c).begin()) it=(c).begin(); it != (c).end(); ++it)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define CLEAR(arr, val) memset(arr, val, sizeof(arr))

#define rep(i, n) for (int i = 0; i < n; ++i)

template <class T> void max_swap(T& a, const T& b) { a = max(a, b); }
template <class T> void min_swap(T& a, const T& b) { a = min(a, b); }

typedef long long ll;
typedef pair<int, int> pint;

const double EPS = 1e-8;
const double PI = acos(-1.0);
const int dx[] = { 0, 1, 0, -1 };
const int dy[] = { 1, 0, -1, 0 };



int main()
{
	int q;
	scanf("%d", &q);
	ll par = 0;
	while (q--)
	{
		int p, n;
		char c;
		scanf("%d %c %d", &p, &c, &n);
		if (c == '(')
			par += n;
		else
			par -= n;

		if (par == 0)
			puts("Yes");
		else
			puts("No");
	}
}