We want to encode a given string $S$ to a binary string. Each alphabet character in $S$ should be mapped to a different variable-length code and the code must not be a prefix of others.

Huffman coding is known as one of ways to obtain a code table for such encoding.

For example, we consider that appearance frequencycies of each alphabet character in $S$ are as follows:

| a| b| c| d| e| f
---|---|---|---|---|---|---
freq.| 45| 13| 12| 16| 9| 5
code| 0| 101| 100| 111| 1101| 1100

We can obtain "Huffman codes" (the third row of the table) using Huffman's algorithm.

The algorithm finds a full binary tree called "Huffman tree" as shown in the following figure.

<image>

To obtain a Huffman tree, for each alphabet character, we first prepare a node which has no parents and a frequency (weight) of the alphabet character. Then, we repeat the following steps:

1. Choose two nodes, $x$ and $y$, which has no parents and the smallest weights.
2. Create a new node $z$ whose weight is the sum of $x$'s and $y$'s.
3. Add edge linking $z$ to $x$ with label $0$ and to $y$ with label $1$. Then $z$ become their parent.
4. If there is only one node without a parent, which is the root, finish the algorithm. Otherwise, go to step 1.



Finally, we can find a "Huffman code" in a path from the root to leaves.

Task

Given a string $S$, output the length of a binary string obtained by using Huffman coding for $S$.

Constraints

* $1 \le |S| \le 10^5$
* $S$ consists of lowercase English letters.

Input


$S$


A string $S$ is given in a line.

Output

Print the length of a binary string in a line.

Examples

Input

abca


Output

6


Input

aaabbcccdeeeffg


Output

41


Input

z


Output

1
#include <bits/stdc++.h>
//typedef
//-------------------------#include <bits/stdc++.h>

const double pi = 3.141592653589793238462643383279;

using namespace std;

//conversion
//------------------------------------------
inline int toInt(string s)
{
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x)
{
  ostringstream sout;
  sout << x;
  return sout.str();
}
inline int readInt()
{
  int x;
  scanf("%d", &x);
  return x;
}

//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;

//container util

//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a) * (a))
#define EACH(i, c) for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

//repetition
//------------------------------------------
#define FOR(i, s, n) for (int i = s; i < (int)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define MOD 1000000007

#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto &a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()

typedef long long ll;
typedef pair<int, int> P;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;

#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)

class UnionFind
{
public:
  vector<int> par;
  vector<int> siz;

  UnionFind(int sz_) : par(sz_), siz(sz_, 1)
  {
    for (ll i = 0; i < sz_; ++i)
      par[i] = i;
  }
  void init(int sz_)
  {
    par.resize(sz_);
    siz.assign(sz_, 1LL);
    for (ll i = 0; i < sz_; ++i)
      par[i] = i;
  }

  int root(int x)
  {
    while (par[x] != x)
    {
      x = par[x] = par[par[x]];
    }
    return x;
  }

  bool merge(int x, int y)
  {
    x = root(x);
    y = root(y);
    if (x == y)
      return false;
    if (siz[x] < siz[y])
      swap(x, y);
    siz[x] += siz[y];
    par[y] = x;
    return true;
  }

  bool issame(int x, int y)
  {
    return root(x) == root(y);
  }

  int size(int x)
  {
    return siz[root(x)];
  }
};

ll modPow(ll x, ll n, ll mod = MOD)
{
  ll res = 1;
  while (n)
  {
    if (n & 1)
      res = (res * x) % mod;

    res %= mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}

#define SIEVE_SIZE 5000000 + 10
bool sieve[SIEVE_SIZE];
void makeSieve()
{
  for (int i = 0; i < SIEVE_SIZE; ++i)
    sieve[i] = true;
  sieve[0] = sieve[1] = false;
  for (int i = 2; i * i < SIEVE_SIZE; ++i)
    if (sieve[i])
      for (int j = 2; i * j < SIEVE_SIZE; ++j)
        sieve[i * j] = false;
}

bool isprime(ll n)
{
  if (n == 0 || n == 1)
    return false;
  for (ll i = 2; i * i <= n; ++i)
    if (n % i == 0)
      return false;
  return true;
}

const int MAX = 2000010;
long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit()
{
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++)
  {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

// 二項係数計算
long long COM(int n, int k)
{
  if (n < k)
    return 0;
  if (n < 0 || k < 0)
    return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long extGCD(long long a, long long b, long long &x, long long &y)
{
  if (b == 0)
  {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extGCD(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
// 負の数にも対応した mod (a = -11 とかでも OK)
inline long long mod(long long a, long long m)
{
  return (a % m + m) % m;
}

// 逆元計算 (ここでは a と m が互いに素であることが必要)
long long modinv(long long a, long long m)
{
  long long x, y;
  extGCD(a, m, x, y);
  return mod(x, m); // 気持ち的には x % m だが、x が負かもしれないので
}
ll GCD(ll a, ll b)
{

  if (b == 0)
    return a;
  return GCD(b, a % b);
}

typedef vector<ll> vec;
typedef vector<vec> mat;

mat mul(mat &A, mat &B)
{
  mat C(A.size(), vec((int)B[0].size()));
  for (int i = 0; i < A.size(); ++i)
  {
    for (int k = 0; k < B.size(); ++k)
    {
      for (int j = 0; j < B[0].size(); ++j)
      {
        C[i][j] = (C[i][j] + A[i][k] % MOD * B[k][j] % MOD) % MOD;
      }
    }
  }
  return C;
}
mat matPow(mat A, ll n)
{
  mat B(A.size(), vec((int)A.size()));

  for (int i = 0; i < A.size(); ++i)
  {
    B[i][i] = 1;
  }

  while (n > 0)
  {
    if (n & 1)
      B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);

  string s;
  map<char, int> mp;
  cin >> s;

  int cnt[26] = {0};
  for (int i = 0; i < s.size(); i++)
    cnt[s[i] - 'a']++;

  priority_queue<int, vector<int>, greater<int>> q;

  for (int i = 0; i < 26; i++)
  {
    if (cnt[i])
      q.push(cnt[i]);
  }

  int ans = 0;
  if (q.size() == 1)
  {
    cout << q.top() << endl;
    return 0;
  }
  while (q.size() > 1)
  {
    int x = q.top();
    q.pop();
    int y = q.top();
    q.pop();

    ans += x + y;
    q.push(x + y);
  }
  cout << ans << endl;

  return 0;
}

