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;
const double eps = 1e-8;
const int N = 1e5 + 7;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
char s[N], z[N];
int g[N], f[N];
int main() {
  scanf("%s", s + 1);
  int l = 0, cnt = 0, top = 0, n = strlen(s + 1), L = 1, R = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '(') {
      z[++top] = '(';
      f[top] = l;
      g[top] = cnt;
      cnt = l = 0;
    } else if (s[i] == '[') {
      z[++top] = '[';
      f[top] = l;
      g[top] = cnt;
      cnt = l = 0;
    } else if (s[i] == ')') {
      if (z[top] != '(') {
        top = cnt = l = 0;
        continue;
      } else {
        cnt += g[top];
        l++;
        l += f[top];
        top--;
        if (cnt > ans) {
          ans = cnt;
          L = i - l * 2 + 1;
          R = i;
        }
      }
    } else {
      if (z[top] != '[') {
        top = cnt = l = 0;
        continue;
      } else {
        cnt++;
        cnt += g[top];
        l++;
        l += f[top];
        top--;
        if (cnt > ans) {
          ans = cnt;
          L = i - l * 2 + 1;
          R = i;
        }
      }
    }
  }
  printf("%d\n", ans);
  for (int i = L; i <= R; i++) putchar(s[i]);
}
