You are given a positive (greater than zero) integer n.

You have to represent n as the sum of integers (possibly negative) consisting only of ones (digits '1'). For example, 24 = 11 + 11 + 1 + 1 and 102 = 111 - 11 + 1 + 1. 

Among all possible representations, you have to find the one that uses the minimum number of ones in total.

Input

The single line contains one integer n (1 ≤ n < 10^{50}).

Output

Print one integer x — the minimum number of ones, such that there exist a representation of n as the sum of integers (possibly negative) that uses x ones in total.

Examples

Input


24


Output


6


Input


102


Output


7
#include <bits/stdc++.h>
using namespace std;
//using namespace atcoder;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define pb push_back
#define fi first
#define se second
//#define int long long
#define INF 1000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void chOut(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using v_bool = vector<bool>;
using v_Pii = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;

const int N = 250;
const int M = 30;
int dp[2][2 * M + 1][N + 1][N + 1];

signed main(){

    string n; cin >> n;
    n = "0" + n;
    int L = SZ(n);

    REP(k, 2 * M + 1) REP(cp, N + 1) REP(cn, N + 1) dp[0][k][cp][cn] = INF;
    dp[0][M][N][N] = 0;
    REP(i, L){
        int d = n[L - 1 - i] - '0';

        REP(k, 2 * M + 1){
            REP(cp, N + 1) IREP(cn, N) chmin(dp[i % 2][k][cp][cn], dp[i % 2][k][cp][cn + 1]);
            REP(cn, N + 1) IREP(cp, N) chmin(dp[i % 2][k][cp][cn], dp[i % 2][k][cp + 1][cn]);
        }
        REP(k, 2 * M + 1) REP(cp, N + 1) REP(cn, N + 1) dp[(i + 1) % 2][k][cp][cn] = INF;

        REP(k0, 2 * M + 1) REP(cp, N + 1) REP(cn, N + 1){
            int t = cp - cn + k0 - M;
            if((t - d) % 10) continue;
            int k = (t - d) / 10 + M;
            chmin(dp[(i + 1) % 2][k][cp][cn], dp[i % 2][k0][cp][cn] + cp + cn);
        }
    }
    int ans = INF;
    REP(cp, N + 1) REP(cn, N + 1) chmin(ans, dp[L % 2][M][cp][cn]);
    Out(ans);

    return 0;
}
