#include <bits/stdc++.h>
using namespace std;
inline void read(int &x) {
  short negative = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') negative = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  x *= negative;
}
inline void write(long long x) {
  static long long sta[35];
  long long top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(sta[--top] + 48);
  puts("");
}
bool prime(int n) {
  if (n == 0 || n == 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int l = -1;
  cout << n << endl;
  for (int i = 0; i < (int)(s.size()); i++) {
    if (i < k) continue;
    char p = s[i - k];
    if (s[i] != p) {
      l = i;
      break;
    }
  }
  if (l == -1) {
    cout << s << endl;
    return 0;
  }
  if (s[l] < s[l - k]) {
    for (int i = l; i < s.size(); i++) {
      s[i] = s[i - k];
    }
    cout << s << endl;
    return 0;
  } else {
    int d = k - 1;
    while (s[d] == '9') {
      s[d] = '0';
      d--;
    }
    s[d]++;
    for (int i = 0; i < (int)(k); i++) {
      for (int j = i; j < n; j += k) s[j] = s[i];
    }
    cout << s << endl;
    return 0;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
inline void read(int &x) {
  short negative = 1;
  x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') negative = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  x *= negative;
}
inline void write(long long x) {
  static long long sta[35];
  long long top = 0;
  do {
    sta[top++] = x % 10, x /= 10;
  } while (x);
  while (top) putchar(sta[--top] + 48);
  puts("");
}
bool prime(int n) {
  if (n == 0 || n == 1) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
int main() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  int l = -1;
  cout << n << endl;
  for (int i = 0; i < (int)(s.size()); i++) {
    if (i < k) continue;
    char p = s[i - k];
    if (s[i] != p) {
      l = i;
      break;
    }
  }
  if (l == -1) {
    cout << s << endl;
    return 0;
  }
  if (s[l] < s[l - k]) {
    for (int i = l; i < s.size(); i++) {
      s[i] = s[i - k];
    }
    cout << s << endl;
    return 0;
  } else {
    int d = k - 1;
    while (s[d] == '9') {
      s[d] = '0';
      d--;
    }
    s[d]++;
    for (int i = 0; i < (int)(k); i++) {
      for (int j = i; j < n; j += k) s[j] = s[i];
    }
    cout << s << endl;
    return 0;
  }
  return 0;
}

```