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 <class T>
inline T _abs(const T n) {
  return (n < 0 ? -n : n);
}
template <class T>
inline T _max(const T a, const T b) {
  return (!(a < b) ? a : b);
}
template <class T>
inline T _min(const T a, const T b) {
  return (a < b ? a : b);
}
template <class T>
inline void _setmax(T &a, const T b) {
  if (a < b) a = b;
}
template <class T>
inline void _setmin(T &a, const T b) {
  if (b < a) a = b;
}
template <class T>
inline T _sqr(const T x) {
  return x * x;
}
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int INF = 0x7f7f7f7f;
const int TMAX = 1 << 18;
string ans;
string s;
int main() {
  cin >> s;
  int i = 0;
  int l = s.length();
  int dot0 = 0, on = 0;
  while (i < l) {
    if (s[i] > 48) {
      on = 1;
    }
    if (s[i] == '.' && !on) {
      dot0 = 1;
      break;
    } else if (s[i] == '.') {
    }
    i++;
  }
  int u = 0;
  while (s[u] == '0' && u < l) {
    u++;
  }
  int pr = 0;
  if (u == l) {
    cout << "0" << endl;
    pr = 1;
  } else {
    if (s[u] == '.') u++;
    while (s[u] == '0' && u < l) {
      u++;
    }
    if (u == l) {
      cout << "0" << endl;
      pr = 1;
    }
  }
  if (!pr) {
    if (dot0) {
      i++;
      int f = 0, coun = 0;
      while (s[i] == 48) {
        i++;
        coun++;
      }
      while (i < l) {
        if (!f) {
          ans.push_back(s[i]);
          ans.push_back('.');
          f = 1;
        } else {
          ans.push_back(s[i]);
        }
        i++;
      }
      i = ans.length();
      while (ans[--i] == '0') {
        ans.erase(i, 1);
      }
      if (ans[i] == '.') {
        ans.erase(i, 1);
      }
      ans.push_back('E');
      ans.push_back('-');
      coun++;
      string g;
      while (coun) {
        char ch = coun % 10 + 48;
        g.push_back(ch);
        coun /= 10;
      }
      int l = g.length();
      while (l) {
        ans.push_back(g[--l]);
      }
      g.clear();
    } else {
      int i = 0, f = 0, coun = 0, f2 = 0;
      while (s[i] == '0') {
        i++;
      }
      while (i < l) {
        if (!f) {
          ans.push_back(s[i]);
          ans.push_back('.');
          f = 1;
        } else if (s[i] != '.') {
          ans.push_back(s[i]);
          if (!f2) {
            coun++;
          }
        } else {
          f2 = 1;
        }
        i++;
      }
      i = ans.length();
      while (ans[--i] == '0') {
        ans.erase(i, 1);
      }
      if (ans[i] == '.') {
        ans.erase(i, 1);
      }
      if (coun) {
        ans.push_back('E');
        string g;
        while (coun) {
          char ch = coun % 10 + 48;
          g.push_back(ch);
          coun /= 10;
        }
        int l = g.length();
        while (l) {
          ans.push_back(g[--l]);
        }
        g.clear();
      }
    }
    cout << ans << endl;
  }
  ans.clear();
}