Problem: An accordion is a string (yes, in the real world accordions are musical instruments, but let's forget about it for a while) which can be represented as a concatenation of: an opening bracket (ASCII code 091), a colon (ASCII code 058), some (possibly zero) vertical line characters (ASCII code 124), another colon, and a closing bracket (ASCII code 093). The length of the accordion is the number of characters in it.

For example, [::], [:||:] and [:|||:] are accordions having length 4, 6 and 7. (:|:), {:||:}, [:], ]:||:[ are not accordions. 

You are given a string s. You want to transform it into an accordion by removing some (possibly zero) characters from it. Note that you may not insert new characters or reorder existing ones. Is it possible to obtain an accordion by removing characters from s, and if so, what is the maximum possible length of the result?

Input

The only line contains one string s (1 ≤ |s| ≤ 500000). It consists of lowercase Latin letters and characters [, ], : and |.

Output

If it is not possible to obtain an accordion by removing some characters from s, print -1. Otherwise print maximum possible length of the resulting accordion.

Examples

Input


|[a:b:|]


Output


4


Input


|]:[|:]


Output


-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
vector<long long> primes;
void sieve(long long maxN) {
  vector<long long> ar(maxN + 1, 0);
  ar[1] = 1;
  for (long long i = 2; i <= maxN; i++)
    if (ar[i] == 0) {
      for (long long j = 2 * i; j <= maxN; j += i) ar[j] = 1;
    }
  for (long long i = 1; i <= maxN; i++)
    if (ar[i] == 0) primes.push_back(i);
}
long long mul(long long a, long long b, long long mod1 = 998244353) {
  long long res = 0;
  a %= mod1;
  while (b) {
    if (b & 1) res = (res + a) % mod1;
    a = (2 * a) % mod1;
    b >>= 1;
  }
  return res;
}
int log_(int a, int b) { return (a > b - 1) ? 1 + log_(a / b, b) : 0; }
void solve() {
  long long n, m, y, q, z, k;
  long long ans = 0;
  long long x;
  int l{-1}, r{-1};
  string s;
  cin >> s;
  n = s.size();
  long long flg = -1;
  long long cnt = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '[') {
      l = i;
      cnt = i;
      break;
    }
  }
  if (l == -1) {
    cout << "-1";
    return;
  };
  for (int i = n - 1; i > l; i--) {
    if (s[i] == ']') {
      r = i;
      break;
    }
  }
  if (r == -1) {
    cout << "-1";
    return;
  }
  cnt += (n - r - 1);
  ;
  int l1{-1}, r1{-1};
  for (int i = l + 1; i < r; i++) {
    if (s[i] == ':') {
      l1 = i;
      break;
    }
  }
  if (l1 == -1) {
    cout << "-1";
    return;
  }
  for (int i = r - 1; i > l; i--) {
    if (s[i] == ':') {
      r1 = i;
      break;
    }
  }
  if (r1 == -1 || l1 == r1) {
    cout << "-1";
    return;
  };
  cnt += (l1 - l - 1) + (r - r1 - 1);
  ;
  for (int i = l1 + 1; i < r1; i++) {
    if (s[i] == '|')
      continue;
    else
      cnt++;
  };
  cout << n - cnt;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, k;
  k = 1;
  {
    solve();
    k++;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  cout << flush;
  return 0;
}