Problem: You are given a positive decimal number x.

Your task is to convert it to the "simple exponential notation".

Let x = a·10b, where 1 ≤ a < 10, then in general case the "simple exponential notation" looks like "aEb". If b equals to zero, the part "Eb" should be skipped. If a is an integer, it should be written without decimal point. Also there should not be extra zeroes in a and b.

Input

The only line contains the positive decimal number x. The length of the line will not exceed 106. Note that you are given too large number, so you can't use standard built-in data types "float", "double" and other.

Output

Print the only line — the "simple exponential notation" of the given number x.

Examples

Input

16


Output

1.6E1


Input

01.23400


Output

1.234


Input

.100


Output

1E-1


Input

100.


Output

1E2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string &s) { return '"' + s + '"'; }
string to_string(char ch) {
  string s;
  s += ch;
  return s;
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}
template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
         to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
string get_stringA(string &a) {
  int n = ((long long)a.size());
  int nonzero_id = n - 1;
  42;
  for (int i = n - 1; i >= 0; i--) {
    42;
    if (a[i] == '.' || a[i] == '0') {
    } else {
      nonzero_id = i;
      break;
    }
  }
  return a.substr(0, nonzero_id + 1);
}
string get_stringB(int b) {
  string ans = "";
  if (b == 0) {
  } else {
    ans += 'E';
    ans += to_string(b);
  }
  return ans;
}
void process_no_dot(string &s, int b, string &ans) {
  int n = ((long long)s.size());
  int nonzero_id = n;
  for (int i = 0; i < n; i++) {
    if (s[i] != '0') {
      nonzero_id = i;
      break;
    }
  }
  assert(nonzero_id < n);
  s = s.substr(nonzero_id, n - nonzero_id);
  n = ((long long)s.size());
  string a = to_string(s[0]) + "." + s.substr(1, n - 1);
  b += n - 1;
  ans = get_stringA(a);
  ans += get_stringB(b);
}
void solve() {
  string s;
  cin >> s;
  int n = ((long long)s.size());
  int dot_id = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == '.') {
      dot_id = i;
    }
  }
  string ans;
  if (dot_id == -1) {
    process_no_dot(s, 0, ans);
  } else {
    string news;
    for (int i = 0; i < n; i++) {
      if (s[i] != '.') {
        news += s[i];
      }
    }
    process_no_dot(news, -(n - 1 - dot_id), ans);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int tc = 1; tc <= t; tc++) {
    solve();
  }
}