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;
using ll = long long;
using ld = long double;
const int M = 11;
int dp[2][M][M][M], t1[M], t2[M][M], t4[M][M][M][M];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  for (int i = 0; i < (M); i++) {
    t1[i] = (i + 1) % 11;
    for (int j = 0; j < (M); j++) {
      t2[i][j] = (j + (i + j) * (j - i + 1 + 11) * 6) % 11;
      for (int k = 0; k < (M); k++)
        for (int d = 0; d < (M); d++)
          t4[i][k][j][d] = (k + (i + j + 10) * (j - i + 11) * 6 + d + 1) % 11;
    }
  }
  ll ans = 0;
  for (int i = 0; i < (n); i++) {
    int d = s[i] - '0', r = i % 2, pr = 1 - r;
    for (int j = 0; j < (M); j++)
      for (int k = 0; k < (M); k++) memset(dp[r][j][k], 0, M * sizeof(int));
    for (int j = 0; j < (M); j++)
      for (int k = 0; k < (M); k++)
        for (int l = d + 1; l < M; l++)
          dp[r][t1[k]][t2[j][k]][t4[j][k][l][d]] += dp[pr][j][k][l];
    if (d > 0) dp[r][1][9][d]++;
    for (int j = 0; j < (M); j++)
      for (int k = 0; k < (M); k++)
        for (int l = 0; l < (M); l++) ans += dp[r][j][k][l];
  }
  cout << ans;
}
