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;
using ll = long long;
const int N = 1e2;
map<string, string> maap;
vector<int> circle;
bool vis[N];
int parent[N];
int chk[N];
int color[N];
int tim[2][N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
map<int, int> mp1;
vector<pair<int, int>> vec;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int cas = 0;
  ll n, m, i, j, cnt1 = 0, cnt2 = 0, cnt = 0, even = 0, odd = 0, len, k, r, l,
                 z = 0, x = 0, y = 0, flag = 0, sum = 0;
  ll a = 0, b = 0, c = 0, d = 0, ans = 0LL, rem, quot, zero = 0, fst = 0,
     null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
     mn = INT_MAX, mx = INT_MIN;
  char ch;
  int h1, h2, m1, m2, h;
  int velo1, velo2, ac1, ac2, time;
  int node, edge, u, v, cost;
  int best, worst;
  double nd, ad, bd, xd;
  string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
  cin >> str;
  stack<char> stk;
  stack<char> stk1;
  stack<char> stk2;
  stack<char> stk3;
  stack<char> stk4;
  n = str.size();
  a = b = c = d = 0;
  for (int i = 0; i < n; i++) {
    if (str[i] == '{' || str[i] == '[' || str[i] == '(' || str[i] == '<') {
      cnt++;
    } else {
      cnt--;
    }
  }
  if (cnt == 0) {
    for (int i = 0; i < n; i++) {
      if (str[i] == '{' || str[i] == '[' || str[i] == '(' || str[i] == '<') {
        stk.push(str[i]);
      } else {
        if (stk.empty()) {
          cout << "Impossible" << endl;
          return 0;
        }
        ch = stk.top();
        if (ch == '(') {
          if (str[i] == ')') {
            stk.pop();
          } else {
            cnt++;
            stk.pop();
          }
        } else if (ch == '{') {
          if (str[i] == '}') {
            stk.pop();
          } else {
            cnt++;
            stk.pop();
          }
        } else if (ch == '[') {
          if (str[i] == ']') {
            stk.pop();
          } else {
            cnt++;
            stk.pop();
          }
        } else if (ch == '<') {
          if (str[i] == '>') {
            stk.pop();
          } else {
            cnt++;
            stk.pop();
          }
        }
      }
    }
    cout << cnt << endl;
  } else {
    cout << "Impossible" << endl;
  }
  return 0;
}