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;
void open() { freopen("data.txt", "r", stdin); }
void out() { freopen("out.txt", "w", stdout); }
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
const int MOD = 1e9 + 7;
int id(char c) {
  if (c == '(') return 1;
  if (c == '[') return 2;
  if (c == '{') return 3;
  if (c == '<') return 4;
  if (c == ')') return 8;
  if (c == ']') return 7;
  if (c == '}') return 6;
  if (c == '>') return 5;
}
char str[maxn];
int main() {
  scanf("%s", str);
  int len = strlen(str);
  int cnt1 = 0, cnt2 = 0;
  for (int i = 0; i < (len); ++i) {
    if (str[i] == '(')
      cnt1++;
    else if (str[i] == '[')
      cnt1++;
    else if (str[i] == '{')
      cnt1++;
    else if (str[i] == '<')
      cnt1++;
    else if (str[i] == ']')
      cnt2++;
    else if (str[i] == '}')
      cnt2++;
    else if (str[i] == ')')
      cnt2++;
    else if (str[i] == '>')
      cnt2++;
  }
  if (cnt1 != cnt2) {
    puts("Impossible");
    return 0;
  }
  bool ok = true;
  int ans = 0;
  stack<int> s;
  for (int i = 0; i < (len); ++i) {
    int tmp = id(str[i]);
    if (s.empty()) {
      s.push(tmp);
      continue;
    }
    if (tmp <= 4) {
      s.push(tmp);
      continue;
    }
    if (tmp + s.top() == 9) {
      s.pop();
      continue;
    } else if (tmp >= 5 && tmp + s.top() != 9) {
      ans++;
      s.pop();
    }
  }
  if (s.size())
    puts("Impossible");
  else
    printf("%d\n", ans);
}