#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
inline void cmin(int &x, int y) {
  if (y < x) x = y;
}
inline void cmax(int &x, int y) {
  if (y > x) x = y;
}
vector<int> v[11];
long long ans, dp[11], pd[11];
void calc() {
  vector<int> vn;
  vn.push_back(0);
  for (int i = 1; i <= 9; i++) vn.push_back(i);
  for (int i = 1; i < 100; i++) {
    int t = i % 11;
    if (vn.size() >= 100) break;
    for (int j = 0; j < t; j++) {
      vn.push_back(vn[i] * 10 + j);
      if (vn[i] <= 10) v[vn[i]].push_back((vn.size() - 1) % 11);
    }
  }
}
int main() {
  calc();
  string s;
  cin >> s;
  ans = 1;
  int val = s[0] - '0';
  dp[val] = 1;
  for (int i = 1; i < s.length(); i++) {
    int val = s[i] - '0';
    for (int j = val + 1; j <= 10; j++) {
      ans += dp[j];
      pd[v[j][val]] += dp[j];
    }
    if (val != 0) {
      pd[val]++;
      ++ans;
    }
    for (int j = 1; j <= 10; j++) {
      dp[j] = pd[j];
      pd[j] = 0;
    }
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Your task is to create 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 MOD = 1e9 + 7;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
inline void cmin(int &x, int y) {
  if (y < x) x = y;
}
inline void cmax(int &x, int y) {
  if (y > x) x = y;
}
vector<int> v[11];
long long ans, dp[11], pd[11];
void calc() {
  vector<int> vn;
  vn.push_back(0);
  for (int i = 1; i <= 9; i++) vn.push_back(i);
  for (int i = 1; i < 100; i++) {
    int t = i % 11;
    if (vn.size() >= 100) break;
    for (int j = 0; j < t; j++) {
      vn.push_back(vn[i] * 10 + j);
      if (vn[i] <= 10) v[vn[i]].push_back((vn.size() - 1) % 11);
    }
  }
}
int main() {
  calc();
  string s;
  cin >> s;
  ans = 1;
  int val = s[0] - '0';
  dp[val] = 1;
  for (int i = 1; i < s.length(); i++) {
    int val = s[i] - '0';
    for (int j = val + 1; j <= 10; j++) {
      ans += dp[j];
      pd[v[j][val]] += dp[j];
    }
    if (val != 0) {
      pd[val]++;
      ++ans;
    }
    for (int j = 1; j <= 10; j++) {
      dp[j] = pd[j];
      pd[j] = 0;
    }
  }
  printf("%I64d", ans);
  return 0;
}

```