#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
char s[N];
struct info {
  char a[12];
  info() { memset(a, 0, sizeof(a)); }
  info operator+(const info &b) const {
    info ans;
    for (int i = (int)(0); i <= (int)(11); i++) ans.a[i] = b.a[a[i]];
    return ans;
  }
} tr[11], st[17][N];
long long ans;
int main() {
  for (int i = (int)(0); i <= (int)(9); i++) {
    for (int j = (int)(0); j <= (int)(11); j++)
      if (j <= i || j == 11)
        tr[i].a[j] = 11;
      else
        tr[i].a[j] = (j * (j - 1) / 2 + i + 10) % 11;
  }
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = (int)(1); i <= (int)(n); i++) {
    s[i] -= '0';
    st[0][i] = tr[s[i]];
  }
  for (int i = (int)(1); i <= (int)(16); i++)
    for (int j = (int)(1); j <= (int)(n - (1 << i) + 1); j++)
      st[i][j] = st[i - 1][j] + st[i - 1][j + (1 << (i - 1))];
  for (int i = (int)(1); i <= (int)(n); i++)
    if (s[i]) {
      info tmp;
      for (int j = (int)(0); j <= (int)(10); j++) tmp.a[j] = j;
      int ri = i + 1;
      for (int j = (int)(16); j >= (int)(0); j--)
        if (ri + (1 << j) - 1 <= n) {
          info ttmp = tmp + st[j][ri];
          if (ttmp.a[s[i]] != 11) tmp = ttmp, ri += 1 << j;
        }
      ans += ri - i;
    }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
char s[N];
struct info {
  char a[12];
  info() { memset(a, 0, sizeof(a)); }
  info operator+(const info &b) const {
    info ans;
    for (int i = (int)(0); i <= (int)(11); i++) ans.a[i] = b.a[a[i]];
    return ans;
  }
} tr[11], st[17][N];
long long ans;
int main() {
  for (int i = (int)(0); i <= (int)(9); i++) {
    for (int j = (int)(0); j <= (int)(11); j++)
      if (j <= i || j == 11)
        tr[i].a[j] = 11;
      else
        tr[i].a[j] = (j * (j - 1) / 2 + i + 10) % 11;
  }
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = (int)(1); i <= (int)(n); i++) {
    s[i] -= '0';
    st[0][i] = tr[s[i]];
  }
  for (int i = (int)(1); i <= (int)(16); i++)
    for (int j = (int)(1); j <= (int)(n - (1 << i) + 1); j++)
      st[i][j] = st[i - 1][j] + st[i - 1][j + (1 << (i - 1))];
  for (int i = (int)(1); i <= (int)(n); i++)
    if (s[i]) {
      info tmp;
      for (int j = (int)(0); j <= (int)(10); j++) tmp.a[j] = j;
      int ri = i + 1;
      for (int j = (int)(16); j >= (int)(0); j--)
        if (ri + (1 << j) - 1 <= n) {
          info ttmp = tmp + st[j][ri];
          if (ttmp.a[s[i]] != 11) tmp = ttmp, ri += 1 << j;
        }
      ans += ri - i;
    }
  printf("%lld\n", ans);
}

```