You are working as an analyst in a company working on a new system for big data storage. This system will store n different objects. Each object should have a unique ID.

To create the system, you choose the parameters of the system — integers m ≥ 1 and b_{1}, b_{2}, …, b_{m}. With these parameters an ID of some object in the system is an array of integers [a_{1}, a_{2}, …, a_{m}] where 1 ≤ a_{i} ≤ b_{i} holds for every 1 ≤ i ≤ m.

Developers say that production costs are proportional to ∑_{i=1}^{m} b_{i}. You are asked to choose parameters m and b_{i} so that the system will be able to assign unique IDs to n different objects and production costs are minimized. Note that you don't have to use all available IDs.

Input

In the only line of input there is one positive integer n. The length of the decimal representation of n is no greater than 1.5 ⋅ 10^{6}. The integer does not contain leading zeros.

Output

Print one number — minimal value of ∑_{i=1}^{m} b_{i}.

Examples

Input

36


Output

10


Input

37


Output

11


Input

12345678901234567890123456789


Output

177
#include <bits/stdc++.h>
using namespace std;
namespace io {
const int L = (1 << 21) + 1;
char ibuf[L], *iS, *iT, obuf[L], *oS = obuf, *oT = obuf + L - 1, c, qu[55];
int f, qr;
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
}
inline void putc(char x) {
  *oS++ = x;
  if (oS == oT) flush();
}
template <class I>
inline void gi(I &x) {
  for (f = 1, c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, L, stdin),
                              (iS == iT ? EOF : *iS++))
                            : *iS++);
       c < '0' || c > '9';
       c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, L, stdin),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    if (c == '-') f = -1;
  for (x = 0; c <= '9' && c >= '0';
       c = (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, L, stdin),
                       (iS == iT ? EOF : *iS++))
                     : *iS++))
    x = x * 10 + (c & 15);
  x *= f;
}
template <class I>
inline void print(I x) {
  if (!x) putc('0');
  if (x < 0) putc('-'), x = -x;
  while (x) qu[++qr] = x % 10 + '0', x /= 10;
  while (qr) putc(qu[qr--]);
}
struct IOC {
  ~IOC() { flush(); }
} _ioc_;
};  // namespace io
using io ::gi;
using io ::print;
using io ::putc;
const int N = 1050005, M = 4200005, L = 1500005, D = 10000;
const int pw10[] = {1, 10, 100, 1000, 10000};
static long double g = log(10) / log(3), PI = acos(-1);
namespace FFT {
struct num {
  double r, i;
  num() {}
  num(double _r, double _i) { r = _r, i = _i; }
  inline friend num operator+(num a, num b) {
    return num(a.r + b.r, a.i + b.i);
  }
  inline friend num operator-(num a, num b) {
    return num(a.r - b.r, a.i - b.i);
  }
  inline friend num operator*(num a, num b) {
    return num(a.r * b.r - a.i * b.i, a.r * b.i + a.i * b.r);
  }
  inline friend num operator*(num a, double b) { return num(a.r * b, a.i * b); }
  inline friend num operator/(num a, double b) { return num(a.r / b, a.i / b); }
  inline num conj() { return num(r, -i); }
  inline num mult() { return num(-i, r); }
};
num pol[M], *ed = pol, *ww[23], *iww[23], *_w;
num *nint(int len) {
  num *r = ed;
  ed += len;
  return r;
}
inline int fxt(int n) {
  int l = 1;
  while (l <= n) l <<= 1;
  return l;
}
void init(int n) {
  int i, j, k;
  for (i = j = 1; j <= n; i++, j <<= 1) {
    ww[i] = nint(j), iww[i] = nint(j);
    for (k = 0; k < j; k++) {
      ww[i][k] = num(cos(PI / j * k), sin(PI / j * k));
      iww[i][k] = ww[i][k].conj();
    }
  }
}
void fft(num a[], int n, int dft) {
  int i, j, k, l, c = 0;
  num u, v;
  for (i = 1, j = n >> 1; i < n - 1; i++) {
    if (i < j) swap(a[i], a[j]);
    for (k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
  }
  for (l = 2, c = 1; l <= n; l <<= 1, c++) {
    if (dft == -1)
      _w = iww[c];
    else
      _w = ww[c];
    for (i = l >> 1, j = 0; j < n; j += l)
      for (k = 0; k < i; k++) {
        u = a[j + k], v = a[j + k + i] * _w[k];
        a[j + k] = u + v, a[j + k + i] = u - v;
      }
  }
  if (dft == -1)
    for (int i = 0; i < n; i++) a[i] = a[i] / n;
}
}  // namespace FFT
struct hugeint {
  vector<long long> a;
  hugeint(int s = 0) { a.push_back(s), mt(); }
  long long &operator[](int i) { return a[i]; }
  int read() {
    static char str[L];
    scanf("%s", str);
    int n = strlen(str), i;
    a.resize((n + 3) >> 2);
    long double cur = 0, e = 1;
    for (i = 0; i < n; i++) {
      a[(n - i - 1) >> 2] += pw10[(n - i - 1) & 3] * (str[i] - '0');
      cur += (str[i] - '0') * e, e *= 0.1;
    }
    if (n == 1 && cur == 1) puts("1"), exit(0);
    cur = log(cur) / log(3) + g * (n - 1);
    mt();
    return cur - 2;
  }
  void mt() {
    int i;
    for (i = 0; i < a.size(); i++)
      if (a[i] >= D) {
        if (i + 1 >= a.size())
          a.push_back(a[i] / D);
        else
          a[i + 1] += a[i] / D;
        a[i] %= D;
      }
    while (a.size() > 1 && *a.rbegin() == 0) a.pop_back();
  }
  hugeint square() {
    using FFT ::fft;
    using FFT ::fxt;
    using FFT ::num;
    static num x[N];
    int i, m = fxt(a.size() << 1);
    memset(x, 0, m * sizeof(num));
    for (i = 0; i < a.size(); i++) x[i] = num(a[i], 0);
    fft(x, m, 1);
    for (i = 0; i < m; i++) x[i] = x[i] * x[i];
    fft(x, m, -1);
    hugeint b;
    b.a.resize(m);
    for (i = 0; i < m; i++) b[i] = x[i].r + 0.5;
    b.mt();
    return b;
  }
  friend hugeint operator*(hugeint a, int b) {
    hugeint ret = a;
    for (long long &v : ret.a) v *= b;
    ret.mt();
    return ret;
  }
  friend bool operator<(hugeint a, hugeint b) {
    if (a.a.size() != b.a.size()) return a.a.size() < b.a.size();
    int l = a.a.size() - 1;
    for (; l >= 0; l--)
      if (a[l] != b[l]) return a[l] < b[l];
    return false;
  }
  void print() {
    int l = a.size() - 1;
    printf("%lld", a[l]);
    for (l--; l >= 0; l--) printf("%4.4lld", a[l]);
    puts("");
  }
} n, m;
int ret, k, dp[2333];
hugeint hugepow(int n) {
  if (n == 0) return hugeint(1);
  hugeint ret = hugepow(n >> 1).square();
  if (n & 1) ret = ret * 3;
  return ret;
}
int main() {
  FFT ::init(1 << 20);
  m = hugepow(ret = max(n.read(), 0)), ret *= 3;
  for (k = 1; m * k < n; k++)
    ;
  int i;
  for (i = 2; i <= k; i++)
    dp[i] = min(dp[(i + 1) / 2] + 2, dp[(i + 2) / 3] + 3);
  printf("%d\n", dp[k] + ret);
  return 0;
}
