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
#include <bits/stdc++.h>
using namespace std;
int v[200002];
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int find(long long int x) {
  if (x == v[x]) return x;
  return v[x] = find(v[x]);
}
bool merge(long long int x, long long int y) {
  long long int a = find(x);
  long long int b = find(y);
  if (a != b) {
    v[a] = b;
    return true;
  }
  return false;
}
void build(vector<long long int> &tree, long long int i, long long int l,
           long long int r, vector<long long int> &v) {
  if (l == r) {
    tree[i] = v[l];
    return;
  }
  long long int mid = (l + r) / 2;
  build(tree, 2 * i, l, mid, v);
  build(tree, 2 * i + 1, mid + 1, r, v);
  tree[i] = tree[2 * i] + tree[2 * i + 1];
}
long long int get(vector<long long int> &tree, long long int i, long long int l,
                  long long int r, long long int ql, long long int qr,
                  vector<long long int> &lazy) {
  if (lazy[i] != -1) {
    if (lazy[i] == 0) {
      tree[i] = 0;
      if (l != r) {
        lazy[2 * i] = 0;
        lazy[2 * i + 1] = 0;
      }
    } else {
      tree[i] = lazy[i];
      long long int mid = (l + r) / 2;
      if (l != r) {
        lazy[2 * i] = mid - l + 1;
        lazy[2 * i + 1] = r - mid;
      }
    }
    lazy[i] = -1;
  }
  if (ql > r || qr < l) return 0;
  if (ql <= l && r <= qr) return (tree[i]);
  long long int mid = (l + r) / 2;
  return (get(tree, 2 * i, l, mid, ql, qr, lazy) +
          get(tree, 2 * i + 1, mid + 1, r, ql, qr, lazy));
}
void update(vector<long long int> &tree, long long int i, long long int l,
            long long int r, long long int ql, long long int qr,
            long long int val, vector<long long int> &lazy) {
  long long int mid = (l + r) / 2;
  if (ql > r || qr < l)
    return;
  else if (ql <= l && r <= qr) {
    tree[i] = (val == 0) ? 0 : r - l + 1;
    if (l != r) {
      lazy[2 * i] = (val == 0) ? 0 : mid - l + 1;
      lazy[2 * i + 1] = (val == 0) ? 0 : r - mid;
    }
    lazy[i] = -1;
  } else {
    update(tree, 2 * i, l, mid, ql, qr, val, lazy);
    update(tree, 2 * i + 1, mid + 1, r, ql, qr, val, lazy);
    tree[i] = tree[2 * i] + tree[2 * i + 1];
  }
}
long long int make(vector<vector<long long int> > &v, long long int a,
                   long long int b, long long int ini, long long int pre,
                   long long int use) {
  if ((ini + a) % 10 == pre || (ini + b) % 10 == pre) {
    v[ini][pre] = 1;
    return 1;
  }
  if (use > 10) return 10000000000001;
  if (v[ini][pre] != 0) {
    return v[ini][pre];
  } else {
    v[ini][pre] = min(make(v, a, b, (ini + a) % 10, pre, use + 1),
                      make(v, a, b, (ini + b) % 10, pre, use + 1)) +
                  1;
    return v[ini][pre];
  }
}
void solve() {
  long long int n, m, p, q;
  string st;
  cin >> st;
  string str = "";
  n = st.length();
  int i = 0;
  while (st[i] == '0') {
    i++;
  }
  while (i != n) {
    str += st[i];
    i++;
  }
  n = str.length();
  int a = n;
  for (long long int i = 0; i < n; i++) {
    if (str[i] == '.') {
      a = i;
    }
  }
  int b = -1;
  for (long long int i = 0; i < n; i++) {
    if (str[i] != '0' && str[i] != '.') {
      b = i;
      break;
    }
  }
  long long int e = a - b - 1;
  if (e < 0) {
    e++;
  }
  string k = "";
  for (int i = b; i < n; i++) {
    if (str[i] != '.') {
      k += str[i];
    }
  }
  int j = k.length() - 1;
  while (k[j] == '0') {
    j--;
  }
  int flag = 0;
  for (int i = 1; i < k.length(); i++) {
    if (k[i] != '0') {
      flag = 1;
    }
  }
  cout << k[0];
  if (flag) {
    cout << ".";
    for (long long int i = 1; i < j + 1; i++) {
      cout << k[i];
    }
  }
  if (e != 0) cout << "E" << e << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int te = 1;
  while (te--) {
    solve();
  }
  return 0;
}
