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).
#include <bits/stdc++.h>
using namespace std;
int nx[6][11][200005];
char inp[200005];
int s[200005];
int c[11][11];
int tt[200005], sm[200005];
int main() {
  for (int j = 0; j < 11; j++)
    for (int i = 0; i < 11; i++)
      if (j == 0)
        c[i][j] = 0;
      else
        c[i][j] = (i + c[(i + 1) % 11][j - 1]) % 11;
  sm[0] = 0, tt[1] = sm[1] = 9;
  for (int i = 2; i < 6; i++) {
    tt[i] = c[(sm[i - 2] + 1) % 11][tt[i - 1]];
    sm[i] = (sm[i - 1] + tt[i]) % 11;
  }
  scanf("%s", inp + 1);
  int n = strlen(inp + 1);
  for (int i = 1; i <= n; i++) s[i] = inp[i] - '0';
  for (int i = n + 1; i >= 1; i--)
    for (int j = 0; j < 11; j++)
      for (int k = 1; k < 6; k++) {
        if (i == n + 1)
          nx[k][j][i] = i;
        else {
          nx[k][j][i] = i;
          if (s[i] < (j + sm[k - 1]) % 11) {
            int ed = (c[(sm[k - 1] + 1) % 11][(j + 10) % 11] + s[i] + 1) % 11;
            int nk = k + 1;
            if (k == 6 - 1) nk -= 2;
            nx[k][j][i] = nx[nk][ed][i + 1];
          }
        }
      }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == 0)
      continue;
    else {
      int ned = nx[1][s[i]][i + 1];
      ans += ned - i;
    }
  }
  cout << ans << endl;
  return 0;
}
