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;
int dp[100100], ans, pos, l, cnt[100100];
char s[100100];
vector<int> v;
int main() {
  scanf("%s", s);
  l = strlen(s);
  for (int i = 0; i <= l; i++) dp[i] = -1;
  v.reserve(100100);
  for (int i = 0; i < l; i++) {
    if (s[i] == '(' || s[i] == '[')
      v.push_back(i);
    else {
      if (v.empty()) continue;
      if ((s[v.back()] == '(' && s[i] == ')') ||
          (s[v.back()] == '[' && s[i] == ']')) {
        dp[v.back()] = i;
        dp[i] = v.back();
        v.pop_back();
      } else {
        v.clear();
      }
    }
  }
  cnt[0] = (s[0] == '[');
  for (int i = 1; i < l; i++) cnt[i] = cnt[i - 1] + (s[i] == '[');
  for (int i = 0; i < l; i++) {
    int x = 0, st = i;
    bool fl = 0;
    while (dp[i] != -1) {
      fl = 1;
      if (i == 0)
        x += cnt[dp[i]];
      else
        x += (cnt[dp[i]] - cnt[i - 1]);
      i = dp[i] + 1;
    }
    if (ans < x) {
      ans = x;
      pos = st;
    }
    if (fl == 1) i--;
  }
  printf("%d\n", ans);
  if (ans > 0) {
    while (dp[pos] != -1) {
      int fin = dp[pos];
      for (int i = pos; i <= fin; i++) printf("%c", s[i]);
      pos = fin + 1;
    }
    printf("\n");
  }
  return 0;
}
