#include <bits/stdc++.h>
using namespace std;
const int N_MAX = 1e5 + 5;
int TABLE[11][11];
int N;
string S;
int dp[11][11][N_MAX];
int how_far(int row, int col, int index) {
  int &answer = dp[row][col][index];
  if (answer >= 0) return answer;
  if (index >= N) return 0;
  row = TABLE[row][col];
  col = S[index] - '0';
  if (row <= col) return 0;
  return answer = 1 + how_far(row, col, index + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int num = 10;
  for (int row = 0; row <= 10; row++)
    for (int col = 0; col < row; col++) {
      TABLE[row][col] = num;
      num = (num + 1) % 11;
    }
  cin >> S;
  N = S.size();
  long long total = N;
  for (int i = 0; i < N; i++)
    if (S[i] == '0') total--;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i + 1 < N; i++) {
    int row = S[i] - '0', col = S[i + 1] - '0';
    if (row <= col) continue;
    total++;
    total += how_far(row, col, i + 2);
  }
  cout << total << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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_MAX = 1e5 + 5;
int TABLE[11][11];
int N;
string S;
int dp[11][11][N_MAX];
int how_far(int row, int col, int index) {
  int &answer = dp[row][col][index];
  if (answer >= 0) return answer;
  if (index >= N) return 0;
  row = TABLE[row][col];
  col = S[index] - '0';
  if (row <= col) return 0;
  return answer = 1 + how_far(row, col, index + 1);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int num = 10;
  for (int row = 0; row <= 10; row++)
    for (int col = 0; col < row; col++) {
      TABLE[row][col] = num;
      num = (num + 1) % 11;
    }
  cin >> S;
  N = S.size();
  long long total = N;
  for (int i = 0; i < N; i++)
    if (S[i] == '0') total--;
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i + 1 < N; i++) {
    int row = S[i] - '0', col = S[i + 1] - '0';
    if (row <= col) continue;
    total++;
    total += how_far(row, col, i + 2);
  }
  cout << total << '\n';
}

```