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;
#pragma comment(linker, "/STACK:102400000,102400000")
const int N = 200005;
const int M = 500005;
const int INF = 0x3f3f3f3f;
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
const double pi = acos(-1.0);
char ch[100005];
int ans[100005];
int main() {
  while (cin >> ch) {
    int n = strlen(ch);
    stack<int> ss;
    for (int i = 0; i < n; i++) {
      if (ss.empty()) {
        ss.push(i);
      } else if (ch[i] == '(' || ch[i] == '[') {
        ss.push(i);
      } else {
        if (ch[i] == ']' && ch[ss.top()] == '[') {
          ss.pop();
        } else if (ch[i] == ')' && ch[ss.top()] == '(') {
          ss.pop();
        } else {
          ss.push(i);
        }
      }
    }
    int k = 0;
    ans[k++] = n;
    while (!ss.empty()) {
      ans[k++] = ss.top();
      ss.pop();
    }
    ans[k] = -1;
    int maxx = 0, l = -1, r = -1;
    for (int i = k; i > 0; i--) {
      int cnt = 0;
      int left = ans[i] + 1, right = ans[i - 1] - 1;
      for (int j = ans[i] + 1; j < ans[i - 1]; j++) {
        if (ch[j] == '[') cnt++;
      }
      if (cnt > maxx) {
        maxx = cnt;
        l = left;
        r = right;
      }
    }
    cout << maxx << endl;
    if (maxx != 0) {
      for (int i = l; i <= r; i++) {
        cout << ch[i];
      }
    }
    cout << endl;
  }
}
