#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int n, k;
string a;
bool IsOK() {
  for (int i = 0; i < k; i++)
    for (int j = i + k; j < n; j += k)
      if (a[j] != a[i]) return false;
  return true;
}
void Propagate() {
  for (int i = 0; i < k; i++)
    for (int j = i + k; j < n; j += k) a[j] = a[i];
}
int IncPos() {
  int p = -1;
  for (int i = 0; i < k; i++)
    if (a[i] < '9') p = i;
  return p;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> a;
  if (IsOK()) {
    cout << n << '\n' << a << '\n';
    return 0;
  }
  string t = a;
  Propagate();
  if (a >= t) {
    cout << n << '\n' << a << '\n';
    return 0;
  }
  int p = IncPos();
  if (p != -1) {
    a[p]++;
    for (int i = p + 1; i < k; i++) a[i] = '0';
  }
  Propagate();
  cout << n << '\n' << a << '\n';
}

### Prompt

Please formulate 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;
using ll = long long;
int n, k;
string a;
bool IsOK() {
  for (int i = 0; i < k; i++)
    for (int j = i + k; j < n; j += k)
      if (a[j] != a[i]) return false;
  return true;
}
void Propagate() {
  for (int i = 0; i < k; i++)
    for (int j = i + k; j < n; j += k) a[j] = a[i];
}
int IncPos() {
  int p = -1;
  for (int i = 0; i < k; i++)
    if (a[i] < '9') p = i;
  return p;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k >> a;
  if (IsOK()) {
    cout << n << '\n' << a << '\n';
    return 0;
  }
  string t = a;
  Propagate();
  if (a >= t) {
    cout << n << '\n' << a << '\n';
    return 0;
  }
  int p = IncPos();
  if (p != -1) {
    a[p]++;
    for (int i = p + 1; i < k; i++) a[i] = '0';
  }
  Propagate();
  cout << n << '\n' << a << '\n';
}

```