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>
#pragma GCC optimize("O3")
using namespace std;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  vector<vector<int>> ch(11, vector<int>(11));
  for (int i = 0; i < 11; ++i) {
    for (int j = 0; j < 11; ++j) {
      for (int k = i; k < j + 11; ++k) ch[i][j] += k % 11;
    }
  }
  vector<vector<vector<vector<int>>>> dp(
      2, vector<vector<vector<int>>>(11,
                                     vector<vector<int>>(11, vector<int>(11))));
  int64_t ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int a = 0; a < 11; ++a) {
      for (int b = 0; b < 11; ++b) {
        for (int c = 0; c < 11; ++c) dp[i & 1 ^ 1][a][b][c] = 0;
      }
    }
    int dig = s[i] - '0';
    for (int a = dig + 1; a < 11; ++a) {
      for (int b = 0; b < 11; ++b) {
        for (int c = 0; c < 11; ++c) {
          if (dp[i & 1][a][b][c] == 0) continue;
          int kv = ch[(b + 1) % 11][a], rv = ch[(b + 1) % 11][(c + 1) % 11];
          int na = (c + kv + dig + 1) % 11;
          int nb = c;
          int nc = (c + rv) % 11;
          dp[i & 1 ^ 1][na][nb][nc] += dp[i & 1][a][b][c];
          ans += dp[i & 1][a][b][c];
        }
      }
    }
    if (dig) {
      dp[i & 1 ^ 1][dig][0][9]++;
      ans++;
    }
  }
  cout << ans << "\n";
}
