Vasya is a regular participant at programming contests and is already experienced in finding important sentences in long statements. Of course, numbers constraints are important — factorization of a number less than 1000000 is easier than of a number less than 1000000000. However, sometimes it's hard to understand the number at the first glance. Could it be shortened? For example, instead of 1000000 you could write 10^{6}, instead of 1000000000 —10^{9}, instead of 1000000007 — 10^{9}+7.

Vasya decided that, to be concise, the notation should follow several rules: 

  * the notation should only consist of numbers, operations of addition ("+"), multiplication ("*") and exponentiation ("^"), in particular, the use of braces is forbidden; 
  * the use of several exponentiation operations in a row is forbidden, for example, writing "2^3^4" is unacceptable; 
  * the value of the resulting expression equals to the initial number; 
  * the notation should consist of the minimal amount of symbols. 



Given n, find the equivalent concise notation for it.

Input

The only line contains a single integer n (1 ≤ n ≤ 10 000 000 000).

Output

Output a concise notation of the number n. If there are several concise notations, output any of them.

Examples

Input

2018


Output

2018


Input

1000000007


Output

10^9+7


Input

10000000000


Output

100^5


Input

2000000000


Output

2*10^9

Note

The third sample allows the answer 10^10 also of the length 5.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  map<long long, string> best;
  vector<set<long long>> vis(12);
  auto get = [&](long long val) {
    if (best.count(val)) return best[val];
    return to_string(val);
  };
  auto relax = [&](long long val, string rep) {
    int len = 0;
    long long tmp = val;
    while (tmp) len++, tmp /= 10;
    len = max(1, len);
    if (val > n or int(rep.size()) >= len) return;
    if (best.count(val) and int(best[val].size()) <= int(rep.size())) return;
    vis[int(best[val].size())].erase(val);
    best[val] = rep;
    vis[int(rep.size())].insert(val);
  };
  for (long long a = 2; a * a <= n; a++) {
    long long p = a * a;
    int b = 2;
    while (p <= n) {
      relax(p, to_string(a) + "^" + to_string(b));
      b++;
      p *= a;
    }
  }
  for (int i = 1; i <= 7; i++) {
    for (int j = i; i + j + 1 <= 7; j++) {
      for (auto x : vis[i]) {
        for (auto y : vis[j]) {
          relax(x * y, get(x) + "*" + get(y));
        }
      }
    }
  }
  for (int i = 1; i <= 5; i++) {
    long long lim = pow(10, 7 - i - 1);
    for (int y = 1; y <= lim; y++) {
      for (auto x : vis[i]) {
        relax(x * y, to_string(y) + "*" + get(x));
      }
    }
  }
  string ans = get(n);
  auto relax2 = [&](string rep) {
    if (int(ans.size()) > int(rep.size())) ans = rep;
  };
  for (long long x = 1; x <= 1e5; x++) {
    relax2(get(n - x) + "+" + get(x));
    if (n % x == 0) relax2(get(n / x) + "*" + get(x));
  }
  for (int i = 1; i <= 4; i++) {
    for (auto x : vis[i]) {
      relax2(get(n - x) + "+" + get(x));
      if (n % x == 0) relax2(get(n / x) + "*" + get(x));
    }
  }
  cout << ans << '\n';
  return (0);
}
