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 int N = int(1e5) + 10;
const int K = 11;
const int MOD = int(1e9) + 7;
const int INF = int(2e9) + 5;
const long long INF64 = 1e18;
int dp[N];
void solve() {
  string s;
  cin >> s;
  stack<int> st;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] == '(' || s[i] == '[' || s[i] == '{') {
      st.push(i);
    } else {
      if (st.empty()) continue;
      int t = st.top();
      if (s[i] == ')' && s[t] == '(') {
        dp[i] = i - t + 1;
        if (t > 0) dp[i] += dp[t - 1];
        st.pop();
      } else if (s[i] == ']' && s[t] == '[') {
        dp[i] = i - t + 1;
        if (t > 0) dp[i] += dp[t - 1];
        st.pop();
      } else if (s[i] == '}' && s[t] == '{') {
        dp[i] = i - t + 1;
        if (t > 0) dp[i] += dp[t - 1];
        st.pop();
      } else {
        while (!st.empty()) st.pop();
      }
    }
  }
  int i = n - 1;
  int mx = 0, index = -1;
  while (i >= 0) {
    if (dp[i] == 0) {
      i--;
      continue;
    }
    int cnt = 0, t = i;
    int x = dp[i];
    for (int j = 0; j < x; j++) {
      if (s[i] == '[') cnt++;
      i--;
    }
    if (cnt > mx) {
      mx = cnt;
      index = t;
    }
  }
  if (mx == 0) {
    cout << 0 << '\n';
    return;
  }
  cout << mx << '\n';
  cout << s.substr(index + 1 - dp[index], dp[index]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
