Traveling around the world you noticed that many shop owners raise prices to inadequate values if the see you are a foreigner.

You define inadequate numbers as follows: 

  * all integers from 1 to 9 are inadequate; 
  * for an integer x ≥ 10 to be inadequate, it is required that the integer ⌊ x / 10 ⌋ is inadequate, but that's not the only condition. Let's sort all the inadequate integers. Let ⌊ x / 10 ⌋ have number k in this order. Then, the integer x is inadequate only if the last digit of x is strictly less than the reminder of division of k by 11. 



Here ⌊ x / 10 ⌋ denotes x/10 rounded down.

Thus, if x is the m-th in increasing order inadequate number, and m gives the remainder c when divided by 11, then integers 10 ⋅ x + 0, 10 ⋅ x + 1 …, 10 ⋅ x + (c - 1) are inadequate, while integers 10 ⋅ x + c, 10 ⋅ x + (c + 1), …, 10 ⋅ x + 9 are not inadequate.

The first several inadequate integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 21, 30, 31, 32 …. After that, since 4 is the fourth inadequate integer, 40, 41, 42, 43 are inadequate, while 44, 45, 46, …, 49 are not inadequate; since 10 is the 10-th inadequate number, integers 100, 101, 102, …, 109 are all inadequate. And since 20 is the 11-th inadequate number, none of 200, 201, 202, …, 209 is inadequate.

You wrote down all the prices you have seen in a trip. Unfortunately, all integers got concatenated in one large digit string s and you lost the bounds between the neighboring integers. You are now interested in the number of substrings of the resulting string that form an inadequate number. If a substring appears more than once at different positions, all its appearances are counted separately.

Input

The only line contains the string s (1 ≤ |s| ≤ 10^5), consisting only of digits. It is guaranteed that the first digit of s is not zero.

Output

In the only line print the number of substrings of s that form an inadequate number.

Examples

Input


4021


Output


6


Input


110


Output


3

Note

In the first example the inadequate numbers in the string are 1, 2, 4, 21, 40, 402. 

In the second example the inadequate numbers in the string are 1 and 10, and 1 appears twice (on the first and on the second positions).
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author lewin
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        DForeigner solver = new DForeigner();
        solver.solve(1, in, out);
        out.close();
    }

    static class DForeigner {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            char[] s = in.next().toCharArray();
            int n = s.length;

            long count = 0;
            int[] reach = new int[11];
            Arrays.fill(reach, n);
            for (int j = s.length - 1; j >= 0; j--) {
                int[] nreach = new int[11];
                for (int pnum = 0; pnum < 11; pnum++) {
                    if (s[j] - '0' >= pnum) {
                        nreach[pnum] = j;
                    } else {
                        int nnum = 10 + s[j] - '0';
                        for (int k = 0; k < pnum; k++) {
                            nnum += k;
                        }
                        nnum %= 11;
                        nreach[pnum] = reach[nnum];
                    }
                }
                if (s[j] != '0') count += reach[s[j] - '0'] - j;
                reach = nreach;
            }
            out.println(count);
//        for (int i = 0; i < s.length; i++) {
//            if (s[i] == '0') continue;
//
//            int cdig = s[i]-'0';
//            int num = cdig;
//            count++;
//            for (int j = i+1; j < s.length; j++) {
//                if (s[j]-'0' >= num) {
//                    break;
//                }
//                count++;
//                int nnum = 10 + s[j]-'0';
//                for (int k = 0; k < num; k++) {
//                    nnum += k;
//                }
//                nnum %= 11;
//                num = nnum;
//            }
//        }
//        out.println(count);
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1 << 16];
        private int curChar;
        private int numChars;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (this.numChars == -1) {
                throw new InputMismatchException();
            } else {
                if (this.curChar >= this.numChars) {
                    this.curChar = 0;

                    try {
                        this.numChars = this.stream.read(this.buf);
                    } catch (IOException var2) {
                        throw new InputMismatchException();
                    }

                    if (this.numChars <= 0) {
                        return -1;
                    }
                }

                return this.buf[this.curChar++];
            }
        }

        public String next() {
            int c;
            while (isSpaceChar(c = this.read())) {
                ;
            }

            StringBuilder result = new StringBuilder();
            result.appendCodePoint(c);

            while (!isSpaceChar(c = this.read())) {
                result.appendCodePoint(c);
            }

            return result.toString();
        }

        public static boolean isSpaceChar(int c) {
            return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void close() {
            writer.close();
        }

        public void println(long i) {
            writer.println(i);
        }

    }
}

