Problem:
Score : 600 points

Problem Statement

A bracket sequence is a string that is one of the following:

An empty string;

The concatenation of (, A, and ) in this order, for some bracket sequence A ;

The concatenation of A and B in this order, for some non-empty bracket sequences A and B /

Given are N strings S_i. Can a bracket sequence be formed by concatenating all the N strings in some order?

Constraints

1 \leq N \leq 10^6

The total length of the strings S_i is at most 10^6.

S_i is a non-empty string consisting of ( and ).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

If a bracket sequence can be formed by concatenating all the N strings in some order, print Yes; otherwise, print No.

Sample Input 1

2
)
(()

Sample Output 1

Yes

Concatenating (() and ) in this order forms a bracket sequence.

Sample Input 2

2
)(
()

Sample Output 2

No

Sample Input 3

4
((()))
((((((
))))))
()()()

Sample Output 3

Yes

Sample Input 4

3
(((
)
)

Sample Output 4

No

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

typedef struct {
	int key, id;
} data;

void merge_sort(data x[], int n)
{
	static data y[1000001] = {};
	if (n <= 1) return;
	merge_sort(&(x[0]), n/2);
	merge_sort(&(x[n/2]), (n+1)/2);
	
	int i, p, q;
	for (i = 0, p = 0, q = n/2; i < n; i++) {
		if (p >= n/2) y[i] = x[q++];
		else if (q >= n) y[i] = x[p++];
		else y[i] = (x[p].key < x[q].key)? x[p++]: x[q++];
	}
	for (i = 0; i < n; i++) x[i] = y[i];
}

typedef struct {
	data obj[1000001];
	int size;
} max_heap;

void push(data x, max_heap* h)
{
	int i = h->size;
	data tmp;
	h->obj[(h->size)++] = x;
	while (1) {
		if (i == 0) break;
		else if (h->obj[i].key > h->obj[(i-1)/2].key) {
			tmp = h->obj[(i-1)/2];
			h->obj[(i-1)/2] = h->obj[i];
			h->obj[i] = tmp;
			i = (i - 1) / 2;
		} else break;
	}
}

data pop(max_heap* h)
{
	int i = 0, j;
	data output = h->obj[0], tmp;
	h->obj[0] = h->obj[--(h->size)];
	while (1) {
		if (i * 2 + 1 >= h->size) break;
		else if (i * 2 + 2 >= h->size) j = i * 2 + 1;
		else if (h->obj[i*2+1].key > h->obj[i*2+2].key) j = i * 2 + 1;
		else j = i * 2 + 2;
		if (h->obj[j].key > h->obj[i].key) {
			tmp = h->obj[j];
			h->obj[j] = h->obj[i];
			h->obj[i] = tmp;
			i = j;
		} else break;
	}
	return output;
}

int main()
{
	int i, j, N;
	char T[1000001];
	data count[1000001];
	scanf("%d", &N);
	for (i = 0; i < N; i++) {
		scanf("%s", T);
		for (j = 0, count[i].key = 0, count[i].id = 0; T[j] != 0; j++) {
			if (T[j] == '(') count[i].id++;
			else count[i].id--;
			if (count[i].id < count[i].key) count[i].key = count[i].id;
		}
	}
	merge_sort(count, N);
	
	int tmp = 0;
	data d;
	max_heap h;
	h.size = 0;
	for (i = N - 1; i >= 0 && count[i].key == 0; i--) tmp += count[i].id;
	for (; i >= 0 && tmp >= -count[i].key; i--) {
		d.key = count[i].id;
		d.id = count[i].key;
		push(d, &h);
	}
	while (h.size > 0) {
		d = pop(&h);
		if (tmp + d.id < 0) break;
		tmp += d.key;
		for (; i >= 0 && tmp >= -count[i].key; i--) {
			d.key = count[i].id;
			push(d, &h);
		}
	}
	if (i < 0 && tmp == 0) printf("Yes\n");
	else printf("No\n");
	fflush(stdout);
	return 0;
}