Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int fx[] = {-1, +0, +1, +0, +1, +1, -1, -1, +0};
int fy[] = {+0, -1, +0, +1, +1, -1, +1, -1, +0};
int day[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
template <typename T>
inline bool isLeap(T y) {
  return (y % 400 == 0) || (y % 100 ? y % 4 == 0 : false);
}
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  if (a < b) swap(a, b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T EGCD(T a, T b, T &x, T &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  T x1, y1;
  T d = EGCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B, T M = MOD) {
  A %= M;
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = MOD) {
  return BigMod(A, M - 2, M);
}
template <typename T>
void Compress(T *in, int n, int f = 0) {
  vector<T> vv;
  for (int i = f; i < n + f; i++) vv.push_back(in[i]);
  sort(vv.begin(), vv.end()), vv.erase(unique(vv.begin(), vv.end()), vv.end());
  for (int i = f; i < n + f; i++)
    in[i] = lower_bound(vv.begin(), vv.end(), in[i]) - vv.begin();
}
template <typename T>
T scani(T &n) {
  n = 0;
  bool negative = false;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') negative = true;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    n = n * 10 + c - 48;
    c = getchar();
  }
  if (negative) n = ~(n - 1);
  return n;
}
template <typename T>
void write(T n, int type = true) {
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  if (!n) {
    putchar('0');
    if (type == 32)
      putchar(' ');
    else if (type)
      putchar('\n');
    return;
  }
  char buff[22];
  int len = 0;
  while (n) buff[len++] = n % 10 + 48, n /= 10;
  for (int i = len - 1; i >= 0; i--) putchar(buff[i]);
  if (type == 32)
    putchar(' ');
  else if (type)
    putchar('\n');
}
int scans(char *a) {
  int i = 0;
  char c = 0;
  while (c < 33) c = getchar();
  while (c > 33) {
    a[i++] = c;
    c = getchar();
  }
  a[i] = 0;
  return i;
}
const int pSz = 2000006;
bool np[pSz + 10];
vector<int> prime;
int prime_size;
void sieve() {
  np[0] = np[1] = 1;
  prime.push_back(2);
  for (long long i = 4; i <= pSz; i += 2) np[i] = 1;
  for (long long i = 3; i <= pSz; i += 2) {
    if (!np[i]) {
      prime.push_back(i);
      for (long long j = i * i; j <= pSz; j += (i << 1)) np[j] = 1;
    }
  }
  prime_size = prime.size();
}
template <typename T>
inline T SqrDis(T x1, T y1, T x2, T y2) {
  return ((x1 - x2) * (x1 - x2)) + ((y1 - y2) * (y1 - y2));
}
template <typename T>
inline T Area2(T Ax, T Ay, T Bx, T By, T Cx, T Cy) {
  T ret = Ax * (By - Cy) + Bx * (Cy - Ay) + Cx * (Ay - By);
  if (ret < 0) return ret = -ret;
  return ret;
}
const int N = 200005;
const int M = 200005;
const unsigned long long hs = 3797;
struct complx {
  long double real, img;
  inline complx() { real = img = 0.0; }
  inline complx(long double x) { real = x, img = 0.0; }
  inline complx(long double x, long double y) { real = x, img = y; }
  inline void operator+=(complx &other) {
    real += other.real, img += other.img;
  }
  inline void operator-=(complx &other) {
    real -= other.real, img -= other.img;
  }
  inline complx operator+(complx &other) {
    return complx(real + other.real, img + other.img);
  }
  inline complx operator-(complx &other) {
    return complx(real - other.real, img - other.img);
  }
  inline complx operator*(complx &other) {
    return complx((real * other.real) - (img * other.img),
                  (real * other.img) + (img * other.real));
  }
};
;
void FFT(vector<complx> &ar, int n, int inv) {
  int i, j, l, len, len2;
  const long double p = 4.0 * inv * acos(0.0);
  for (i = 1, j = 0; i < n; i++) {
    for (l = n >> 1; j >= l; l >>= 1) j -= l;
    j += l;
    if (i < j) swap(ar[i], ar[j]);
  }
  for (len = 2; len <= n; len <<= 1) {
    long double ang = 2 * acos(-1.0) / len * inv;
    complx wlen(cos(ang), sin(ang));
    for (i = 0; i < n; i += len) {
      complx w(1);
      for (j = 0; j < len / 2; j++) {
        complx u = ar[i + j];
        complx v = ar[i + j + len / 2] * w;
        ar[i + j] = u + v;
        ar[i + j + len / 2] = u - v;
        w = w * wlen;
      }
    }
  }
  if (inv == -1) {
    long double tmp = 1.0 / n;
    for (i = 0; i < n; i++) ar[i].real *= tmp;
  }
}
vector<complx> Mul(const vector<complx> &x, const vector<complx> &y) {
  int n = 1;
  while (n <= x.size() + y.size()) n = n * 2;
  vector<complx> A(n), B(n);
  for (int i = (0); i < (x.size()); ++i) A[i] = x[i];
  for (int i = (0); i < (y.size()); ++i) B[i] = y[i];
  FFT(A, n, 1);
  FFT(B, n, 1);
  for (int i = (0); i < (n); ++i) A[i] = A[i] * B[i];
  FFT(A, n, -1);
  return A;
}
int in[N];
long long res[N];
string num;
vector<long long> numBs;
long long maxPower3, base;
vector<long long> bruteForceMul(vector<long long> &A, vector<long long> &B,
                                long long bs = 10) {
  vector<long long> a = A, b = B;
  if (a.size() > b.size()) swap(a, b);
  vector<long long> r;
  long long carry = 0;
  long long m = a[0];
  for (int i = b.size() - 1, j = 0; i >= 0; i--) {
    long long cur = (carry + b[i] * m);
    r.push_back(cur % bs);
    carry = cur / bs;
  }
  if (carry) r.push_back(carry);
  reverse(r.begin(), r.end());
  return r;
}
vector<long long> bigNumMul(vector<long long> &a, vector<long long> &b,
                            long long bs = 10) {
  if (min(b.size(), a.size()) == 1) return bruteForceMul(a, b, bs);
  string r;
  int na = a.size();
  int nb = b.size();
  vector<complx> A(na), B(nb);
  for (int i = 0; i < na; i++) A[i] = complx(a[na - i - 1]);
  for (int i = 0; i < nb; i++) B[i] = complx(b[nb - i - 1]);
  int n = na + nb - 1;
  vector<complx> R = Mul(A, B);
  vector<long long> rs(n + 1, 0);
  for (int i = 0; i < n; i++) rs[i] = round(R[i].real);
  for (int i = 0; i < n; i++) {
    rs[i + 1] += rs[i] / bs;
    rs[i] %= bs;
  }
  while (n > 0 && rs[n] == 0) n--;
  n++;
  rs.resize(n);
  reverse(rs.begin(), rs.end());
  return rs;
}
char bigNumCompare(vector<long long> &a, vector<long long> &b) {
  if (a.size() == b.size()) {
    for (int i = 0; i < a.size(); i++) {
      if (a[i] < b[i]) return '<';
      if (a[i] > b[i]) return '>';
    }
    return '=';
  }
  return (a.size() < b.size()) ? '<' : '>';
}
vector<long long> bigPow(vector<long long> R, vector<long long> a, int b) {
  while (b) {
    if (b & 1) R = bigNumMul(R, a, base);
    a = bigNumMul(a, a, base);
    b = b / 2;
  }
  return R;
}
int ans_me() {
  int L = numBs.size();
  int P = (num.size() - 1) * log(10) / log(3);
  P = max(0, P - 1);
  int sum = 3 * P;
  int ret = 0x3f3f3f3f;
  vector<long long> temp;
  vector<long long> b = vector<long long>{3};
  vector<long long> b2 = vector<long long>{2};
  vector<long long> b4 = vector<long long>{4};
  vector<long long> threePowP = bigPow(vector<long long>{1}, b, P);
  while (bigNumCompare(threePowP, numBs) == '<') {
    if (bigNumCompare(temp = bigNumMul(threePowP, b2, base), numBs) != '<')
      ret = min(ret, sum + 2);
    if (bigNumCompare(temp = bigNumMul(threePowP, b4, base), numBs) != '<')
      ret = min(ret, sum + 4);
    threePowP = bigNumMul(threePowP, b, base);
    sum += 3;
  }
  ret = min(ret, sum);
  return ret;
}
vector<long long> convertBase(string nc, long long bs) {
  long long pw = 0, sv = bs;
  vector<long long> ret;
  while (sv % 10 == 0) pw++, sv /= 10;
  long long r = nc.size() % pw;
  r = (pw - r) % pw;
  while (r--) nc = '0' + nc;
  for (int i = 0; i < nc.size() / pw; i++) {
    int c = 0;
    for (int j = 0; j < pw; j++) c = c * 10 + nc[i * pw + j] - '0';
    ret.push_back(c);
  }
  return ret;
}
int main() {
  base = 10000;
  maxPower3 = 1500000 * log(10) / log(3);
  maxPower3++;
  cerr << maxPower3 << "\n";
  cin >> num;
  numBs = convertBase(num, base);
  int res = ans_me();
  res = max(res, 1);
  cout << res << "\n";
  return 0;
}