#include <bits/stdc++.h>
using namespace std;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
int main() {
  int N, K;
  cin >> N >> K;
  string A;
  cin >> A;
  string ret = A;
  for (int i = (0), i_end_ = (ret.length()); i < i_end_; i++) ret[i] = A[i % K];
  if (ret < A) {
    bool carry = true;
    int k = K - 1;
    while (carry) {
      carry = (ret[k] == '9');
      if (carry)
        ret[k] = '0';
      else
        ret[k]++;
      k--;
    }
    for (int i = (0), i_end_ = (ret.length()); i < i_end_; i++)
      ret[i] = ret[i % K];
  }
  cout << ret.length() << endl;
  cout << ret << endl;
}

### Prompt

Generate a Cpp solution to the following problem:

You are given an integer x of n digits a_1, a_2, …, a_n, which make up its decimal notation in order from left to right.

Also, you are given a positive integer k < n.

Let's call integer b_1, b_2, …, b_m beautiful if b_i = b_{i+k} for each i, such that 1 ≤ i ≤ m - k.

You need to find the smallest beautiful integer y, such that y ≥ x. 

Input

The first line of input contains two integers n, k (2 ≤ n ≤ 200 000, 1 ≤ k < n): the number of digits in x and k.

The next line of input contains n digits a_1, a_2, …, a_n (a_1 ≠ 0, 0 ≤ a_i ≤ 9): digits of x.

Output

In the first line print one integer m: the number of digits in y.

In the next line print m digits b_1, b_2, …, b_m (b_1 ≠ 0, 0 ≤ b_i ≤ 9): digits of y.

Examples

Input


3 2
353


Output


3
353


Input


4 2
1234


Output


4
1313

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
int main() {
  int N, K;
  cin >> N >> K;
  string A;
  cin >> A;
  string ret = A;
  for (int i = (0), i_end_ = (ret.length()); i < i_end_; i++) ret[i] = A[i % K];
  if (ret < A) {
    bool carry = true;
    int k = K - 1;
    while (carry) {
      carry = (ret[k] == '9');
      if (carry)
        ret[k] = '0';
      else
        ret[k]++;
      k--;
    }
    for (int i = (0), i_end_ = (ret.length()); i < i_end_; i++)
      ret[i] = ret[i % K];
  }
  cout << ret.length() << endl;
  cout << ret << endl;
}

```