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;
long long inline gcd(long long x, long long y) {
  return !y ? (long long)abs(x) : gcd(y, x % y);
}
long long inline lcm(long long a, long long b) {
  return ((long long)abs((a / gcd(a, b)) * b));
}
double inline cartesian_dist(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long bigmod(long long b, long long p, long long m) {
  if (!p)
    return 1 % m;
  else if (p & 1)
    return (((b % m) + m) % m * bigmod(b, p - 1, m)) % m;
  else {
    long long ret = bigmod(b, p / 2, m);
    return (ret * ret) % m;
  }
}
long long inverse(long long a, long long m) { return bigmod(a, m - 2, m); }
long long extended(long long a, long long m) {
  long long oldr = a, r = m, oldans = 1, ans = 0, quotient;
  while (r) {
    quotient = oldr / r;
    oldr = oldr - r * quotient;
    swap(oldr, r);
    oldans = oldans - ans * quotient;
    swap(oldans, ans);
  }
  return ((oldans % m) + m) % m;
}
const double PI = acos(-1.0);
const double eps = 10E-12;
string s, pre, post;
int main() {
  cin >> s;
  int idx = 0;
  while (s[idx] != '.' && idx < s.size()) pre += s[idx++];
  ++idx;
  while (idx < s.size()) post += s[idx++];
  reverse(pre.begin(), pre.end());
  while (pre.size() && pre[pre.size() - 1] == '0') pre.erase(pre.size() - 1);
  while (post.size() && post[post.size() - 1] == '0')
    post.erase(post.size() - 1);
  reverse(pre.begin(), pre.end());
  if (!pre.size() && !post.size()) {
    printf("0\n");
    return 0;
  }
  if (pre.size()) {
    int exp = pre.size() - 1;
    if (post.size()) {
      putchar(pre[0]);
      putchar('.');
      for (int i = 1; i < pre.size(); i++) putchar(pre[i]);
      cout << post;
      if (exp) printf("E%d", exp);
      printf("\n");
      return 0;
    }
    while (pre.size() && pre[pre.size() - 1] == '0') pre.erase(pre.size() - 1);
    putchar(pre[0]);
    if (pre.size() > 1) {
      putchar('.');
      for (int i = 1; i < pre.size(); i++) putchar(pre[i]);
    }
    if (exp) printf("E%d", exp);
    printf("\n");
    return 0;
  }
  int exp = 1;
  reverse(post.begin(), post.end());
  while (post.size() && post[post.size() - 1] == '0') {
    post.erase(post.size() - 1);
    exp++;
  }
  exp *= -1;
  reverse(post.begin(), post.end());
  putchar(post[0]);
  if (post.size() > 1) {
    putchar('.');
    for (int i = 1; i < post.size(); i++) putchar(post[i]);
  }
  if (exp) printf("E%d", exp);
  printf("\n");
  return 0;
}