Problem: Suppose you have a special x-y-counter. This counter can store some value as a decimal number; at first, the counter has value 0. 

The counter performs the following algorithm: it prints its lowest digit and, after that, adds either x or y to its value. So all sequences this counter generates are starting from 0. For example, a 4-2-counter can act as follows:

  1. it prints 0, and adds 4 to its value, so the current value is 4, and the output is 0; 
  2. it prints 4, and adds 4 to its value, so the current value is 8, and the output is 04; 
  3. it prints 8, and adds 4 to its value, so the current value is 12, and the output is 048; 
  4. it prints 2, and adds 2 to its value, so the current value is 14, and the output is 0482; 
  5. it prints 4, and adds 4 to its value, so the current value is 18, and the output is 04824. 



This is only one of the possible outputs; for example, the same counter could generate 0246802468024 as the output, if we chose to add 2 during each step.

You wrote down a printed sequence from one of such x-y-counters. But the sequence was corrupted and several elements from the sequence could be erased.

Now you'd like to recover data you've lost, but you don't even know the type of the counter you used. You have a decimal string s — the remaining data of the sequence. 

For all 0 ≤ x, y < 10, calculate the minimum number of digits you have to insert in the string s to make it a possible output of the x-y-counter. Note that you can't change the order of digits in string s or erase any of them; only insertions are allowed.

Input

The first line contains a single string s (1 ≤ |s| ≤ 2 ⋅ 10^6, s_i ∈ \{0 - 9\}) — the remaining data you have. It's guaranteed that s_1 = 0.

Output

Print a 10 × 10 matrix, where the j-th integer (0-indexed) on the i-th line (0-indexed too) is equal to the minimum number of digits you have to insert in the string s to make it a possible output of the i-j-counter, or -1 if there is no way to do so.

Example

Input


0840


Output


-1 17 7 7 7 -1 2 17 2 7 
17 17 7 5 5 5 2 7 2 7 
7 7 7 4 3 7 1 7 2 5 
7 5 4 7 3 3 2 5 2 3 
7 5 3 3 7 7 1 7 2 7 
-1 5 7 3 7 -1 2 9 2 7 
2 2 1 2 1 2 2 2 0 1 
17 7 7 5 7 9 2 17 2 3 
2 2 2 2 2 2 0 2 2 2 
7 7 5 3 7 7 1 3 2 7 

Note

Let's take, for example, 4-3-counter. One of the possible outcomes the counter could print is 0(4)8(1)4(7)0 (lost elements are in the brackets).

One of the possible outcomes a 2-3-counter could print is 0(35)8(1)4(7)0.

The 6-8-counter could print exactly the string 0840.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class CF_EDU70_B {

    int dp[] = new int[10];

    void solve() {
        String a = IO.next();
        int occ[]= new int[10];
        for(int i=1; i<a.length();i++)
            occ[(a.charAt(i)-a.charAt(i-1)+10)%10]++;


        for(int x=0; x<=9; x++) {
            for (int y = 0; y <= 9; y++) {
                int sum=0;
                for (int p = 0; p <= 9; p++) {
                    Arrays.fill(dp, 0);
                    int ans = f(p, x, y);
                    Arrays.fill(dp, 0);
                    ans=Math.min(ans, f(p,y,x));
                    if(ans>90 && occ[p]>0) {
                        sum = -1;
                        break;
                    }
                    sum+=occ[p] *ans;
                }
                if(sum>=0)
                    sum-=a.length()-1;
                System.out.print(sum+" ");
            }
            System.out.println();
        }
    }

    int f(int diff, int x, int y) {
        if(diff<0)
            diff+=10;
        if(diff==x || diff==y)
            return 1;
        if (dp[diff]>0)
            return dp[diff];
        dp[diff]=99;
        return dp[diff]=1+Math.min(f(diff-x,x,y), f(diff-y,x,y));
    }

    //*************
    //*************
    //TEMPLATE CODE

    static long MOD = 0;

    public static void main(String[] args) {
        CF_EDU70_B instance = new CF_EDU70_B();
        int testcases = 1;//IO.nextInt();
        while(testcases-->0)
            instance.solve();
    }

    static class IO {
        private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        private static StringTokenizer st;

        static String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        static int nextInt() {
            return Integer.parseInt(next());
        }

        static long nextLong() {
            return Long.parseLong(next());
        }

        static double nextDouble() {
            return Double.parseDouble(next());
        }

        static int[] nextIntArray(int size) {
            int a[] = new int[size];
            for (int i = 0; i < size; i++)
                a[i] = nextInt();
            return a;
        }

        static long[] nextLongArray(int size) {
            long a[] = new long[size];
            for (int i = 0; i < size; i++)
                a[i] = nextLong();
            return a;
        }

        static double[] nextDoubleArray(int size) {
            double a[] = new double[size];
            for (int i = 0; i < size; i++)
                a[i] = nextDouble();
            return a;
        }
    }
}