Problem: You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

Input

[&lt;}){}


Output

2

Input

{()}[]


Output

0

Input

]]


Output

Impossible

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool CMP(int a, int b) { return a > b; }
long long int p = 1e9 + 7;
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int cnt = 0, cnt1 = 0, u;
vector<vector<long long int>> adj1(200010);
vector<long long int> adj;
long long int vis[310068], last[10000], last1[10000];
vector<long long int> viso;
bool ok = false;
long long int dp[200010];
bool mycompare(string s, string t) {
  string x = s + t;
  string y = t + s;
  if (x > y)
    return 1;
  else
    return 0;
}
vector<string> v1;
long long int maxi = 1e7;
set<long long int> w;
int main() {
  string s;
  cin >> s;
  long long int n = s.length(), i, ans = 0;
  stack<char> q;
  for (i = 0; i < n; i++) {
    if (s[i] == '{' || s[i] == '<' || s[i] == '[' || s[i] == '(')
      q.push(s[i]);
    else {
      if (q.empty()) {
        cout << "Impossible" << endl;
        return 0;
      }
      char p = q.top();
      if (p == '{') {
        if (s[i] != '}') ans++;
      } else if (p == '[') {
        if (s[i] != ']') ans++;
      } else if (p == '<') {
        if (s[i] != '>') ans++;
      } else if (p == '(') {
        if (s[i] != ')') ans++;
      }
      q.pop();
    }
  }
  if (!q.empty())
    cout << "Impossible" << endl;
  else
    cout << ans << endl;
}