#include <bits/stdc++.h>
using namespace std;
vector<long long int> fact(100005);
vector<long long int> prime_check(1e6 + 2);
vector<long long int> primes;
long long powmod(long long x, long long y, long long m) {
  long long res = 1LL;
  while (y) {
    if (y & 1) res = (res * x) % m;
    x = (x * x) % m;
    y /= 2;
  }
  return res;
}
void init_fact() {
  fact[0] = 1;
  for (long long int i = 1; i < 100005; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000009;
  }
}
long long int C(long long int n, long long int r) {
  if (n < r) return 0;
  long long int temp =
      fact[n] * powmod(fact[r], 1000000009 - 2, 1000000009) % 1000000009;
  temp *= powmod(fact[n - r], 1000000009 - 2, 1000000009);
  temp %= 1000000009;
  return temp;
}
void Sieve() {
  for (long long int i = 2; i <= 1e6 + 1; i++) {
    if (prime_check[i]) continue;
    primes.push_back(i);
    for (long long int j = 2 * i; j <= 1e6 + 2; j += i) prime_check[j] = i;
  }
}
long long int phi(long long int n) {
  long long int i, res = n;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      res -= res / i;
    }
  if (n > 1) res -= res / n;
  return res;
}
long long int n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  string s;
  cin >> s;
  string t;
  t = s.substr(0, k);
  long long int pos = 1;
  for (long long int i = 0; i < n; i++) {
    if (s[i] < t[i % k])
      s[i] = t[i % k], pos = 2;
    else if (s[i] > t[i % k] && pos == 1) {
      pos = 0;
      break;
    }
  }
  if (pos == 0) {
    for (long long int i = k - 1; i >= 0; i--) {
      if (t[i] == '9')
        t[i] = '0';
      else {
        t[i]++;
        break;
      }
    }
  }
  cout << n << "\n";
  for (long long int i = 0; i < n; i++) {
    cout << t[i % k];
  }
  cout << "\n";
}

### Prompt

Your task is to create 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;
vector<long long int> fact(100005);
vector<long long int> prime_check(1e6 + 2);
vector<long long int> primes;
long long powmod(long long x, long long y, long long m) {
  long long res = 1LL;
  while (y) {
    if (y & 1) res = (res * x) % m;
    x = (x * x) % m;
    y /= 2;
  }
  return res;
}
void init_fact() {
  fact[0] = 1;
  for (long long int i = 1; i < 100005; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000009;
  }
}
long long int C(long long int n, long long int r) {
  if (n < r) return 0;
  long long int temp =
      fact[n] * powmod(fact[r], 1000000009 - 2, 1000000009) % 1000000009;
  temp *= powmod(fact[n - r], 1000000009 - 2, 1000000009);
  temp %= 1000000009;
  return temp;
}
void Sieve() {
  for (long long int i = 2; i <= 1e6 + 1; i++) {
    if (prime_check[i]) continue;
    primes.push_back(i);
    for (long long int j = 2 * i; j <= 1e6 + 2; j += i) prime_check[j] = i;
  }
}
long long int phi(long long int n) {
  long long int i, res = n;
  for (i = 2; i * i <= n; i++)
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      res -= res / i;
    }
  if (n > 1) res -= res / n;
  return res;
}
long long int n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  string s;
  cin >> s;
  string t;
  t = s.substr(0, k);
  long long int pos = 1;
  for (long long int i = 0; i < n; i++) {
    if (s[i] < t[i % k])
      s[i] = t[i % k], pos = 2;
    else if (s[i] > t[i % k] && pos == 1) {
      pos = 0;
      break;
    }
  }
  if (pos == 0) {
    for (long long int i = k - 1; i >= 0; i--) {
      if (t[i] == '9')
        t[i] = '0';
      else {
        t[i]++;
        break;
      }
    }
  }
  cout << n << "\n";
  for (long long int i = 0; i < n; i++) {
    cout << t[i % k];
  }
  cout << "\n";
}

```