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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class B {
	static FastReader scan;
	static PrintWriter out;

	public static void main(String[] args) throws FileNotFoundException {
		Solver solver = new Solver();
		scan = new FastReader();
		out = new PrintWriter(System.out);
		int testCases = 1;
		for (int i = 1; i <= testCases; i++) {
			// out.print("Case #" + i + ": ");
			solver.solve();
		}
		out.close();
	}

	static class Solver {
		static String a;
		static int max = 100;
		static int[][][] dp = new int[max][max][max];
		void solve() {
			a = scan.next();
			for (int i = 0; i < 11; i++) {
				for (int j = 0; j < 11; j++) {
					Arrays.fill(dp[i][j], Integer.MAX_VALUE);
				}
			}
			dp[0][0][0] = 0;
			for (int j = 0; j <= 10; j++) {
				for (int k = 0; k <= 10; k++) {
					for (int jj = 0; jj <= 100; jj++) {
						for (int kk = 0; kk <= 100; kk++) {
							if(jj == 0 && kk == 0) continue;
							int at = (j*jj+k*kk)%10;
							dp[j][k][at] = Math.min(dp[j][k][at], jj+kk);
						}
					}
				}
			}
			int[][] store = new int[10][10];
			for(int i = 0; i < 10; i++) {
				for(int j = i; j < 10; j++) {
					store[i][j] = store[j][i] = calc(i,j);
				}
			}
			for (int i = 0; i < 10; i++) {
				for (int j = 0; j < 10; j++) {
					out.print(store[i][j] + " ");
				}
				out.println();
			}
			
		}

		static int calc(int i, int j) {
			int total = 0;
			for (int k = 1; k < a.length(); k++) {
				int curr = a.charAt(k)-'0', prev = a.charAt(k-1)-'0';
				int diff = prev > curr ? curr + 10 - prev : curr - prev;
				if(dp[i][j][diff] == Integer.MAX_VALUE) return -1;
				total += Math.max(0, dp[i][j][diff]-1);
			}
			return total;
		}

	}

	// Sathvik's Template Stuff BELOW!!!!!!!!!!!!!!!!!!!!!!

	static class DSU {
		int[] root, size;
		int n;

		DSU(int n) {
			this.n = n;
			root = new int[n];
			size = new int[n];
			for (int i = 0; i < n; i++) {
				root[i] = i;
				size[i] = 1;
			}
		}

		int findParent(int idx) {
			while (root[idx] != idx) {
				root[idx] = root[root[idx]];
				idx = root[idx];
			}
			return idx;
		}

		boolean union(int x, int y) {
			int parX = findParent(x);
			int parY = findParent(y);
			if (parX == parY)
				return false;
			if (size[parX] < size[parY]) {
				root[parY] = parX;
				size[parX] += size[parY];
			} else {
				root[parX] = parY;
				size[parY] += size[parX];
			}
			return true;
		}
	}

	static class Extra {
		static void sort(int[] a) {
			Integer[] aa = new Integer[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(long[] a) {
			Long[] aa = new Long[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(double[] a) {
			Double[] aa = new Double[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static void sort(char[] a) {
			Character[] aa = new Character[a.length];
			for (int i = 0; i < aa.length; i++)
				aa[i] = a[i];
			Arrays.sort(aa);
			for (int i = 0; i < aa.length; i++)
				a[i] = aa[i];
		}

		static long gcd(long a, long b) {
			while (b > 0) {
				long temp = b;
				b = a % b;
				a = temp;
			}
			return a;
		}

		static long lcm(long a, long b) {
			return a * (b / gcd(a, b));
		}

		static boolean isPrime(long n) {
			if (n <= 1)
				return false;
			if (n <= 3)
				return true;
			if (n % 2 == 0 || n % 3 == 0)
				return false;
			for (long i = 5; i * i <= n; i = i + 6) {
				if (n % i == 0 || n % (i + 2) == 0)
					return false;
			}
			return true;
		}
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

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

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

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

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}