A bracket sequence is a string, containing only characters "(", ")", "[" and "]".

A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()[]", "([])" are correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.

A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.

You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.

Input

The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.

Output

In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.

Examples

Input

([])


Output

1
([])


Input

(((


Output

0
#include <bits/stdc++.h>
using namespace std;
char s[110000];
int dp[110000];
int sum[110000];
int get(int l, int r) { return sum[r] - sum[l]; }
bool match(char c1, char c2) {
  return (c1 == '(' && c2 == ')') || (c1 == '[' && c2 == ']');
}
int main() {
  gets(s);
  int len = strlen(s);
  for (int i = 0; i < len; i++) sum[i + 1] = sum[i] + (s[i] == '[');
  int best = 0;
  int bestl = 0, bestr = 0;
  dp[len - 1] = dp[len] = 0;
  for (int i = len - 2; i >= 0; i--) {
    if (!match(s[i], s[i + dp[i + 1] + 1])) continue;
    dp[i] = dp[i + 1] + 2 + dp[i + dp[i + 1] + 2];
    if (get(i, i + dp[i]) > best) {
      best = get(i, i + dp[i]);
      bestl = i;
      bestr = i + dp[i];
    }
  }
  cout << best << endl;
  cout << string(s + bestl, s + bestr) << endl;
  return 0;
}
