#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
unordered_map<long long, string> q;
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 2; i <= 100000; i++) {
    for (long long j = i * i, k = 2; j <= n; j *= i, k++) {
      string s = to_string(i) + "^" + to_string(k);
      if (!q.count(j)) q[j] = to_string(j);
      if (s.size() < q[j].size()) q[j] = s;
    }
  }
  string best = to_string(n);
  for (auto p : q) {
    string now;
    if (n / p.first > 1)
      now += (q.count(n / p.first) ? q[n / p.first] : to_string(n / p.first)) +
             "*";
    now += p.second;
    if (n % p.first)
      now += "+" +
             (q.count(n % p.first) ? q[n % p.first] : to_string(n % p.first));
    if (now.size() < best.size()) best = now;
  }
  cout << best << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
unordered_map<long long, string> q;
int main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 2; i <= 100000; i++) {
    for (long long j = i * i, k = 2; j <= n; j *= i, k++) {
      string s = to_string(i) + "^" + to_string(k);
      if (!q.count(j)) q[j] = to_string(j);
      if (s.size() < q[j].size()) q[j] = s;
    }
  }
  string best = to_string(n);
  for (auto p : q) {
    string now;
    if (n / p.first > 1)
      now += (q.count(n / p.first) ? q[n / p.first] : to_string(n / p.first)) +
             "*";
    now += p.second;
    if (n % p.first)
      now += "+" +
             (q.count(n % p.first) ? q[n % p.first] : to_string(n % p.first));
    if (now.size() < best.size()) best = now;
  }
  cout << best << endl;
  return 0;
}

```