#include <bits/stdc++.h>
using namespace std;
const long long int MODA = 1000000007;
vector<long long int> primefactors;
vector<long long int> factors;
long long int fact[1025] = {};
bool isPowerTwo(long long int x) { return (x && !(x & (x - 1))); }
long long int modmul(long long int a, long long int b) {
  return ((a % MODA) * (b % MODA)) % MODA;
}
long long int modadd(long long int a, long long int b) {
  return ((a % MODA) + (b % MODA) + MODA) % MODA;
}
long long int modsub(long long int a, long long int b) {
  return ((a % MODA) - (b % MODA) + MODA) % MODA;
}
bool isSubstring(string s1, string s2) {
  if (s1.find(s2) != string::npos)
    return true;
  else
    return false;
}
void generateFactorial(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) fact[i] = (i * 1ll * fact[i - 1]);
}
void generateFactorialMOD(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * 1ll * fact[i - 1]) % MODA;
}
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
long long int powermod(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MODA;
    a = (a * a) % MODA;
    b >>= 1;
  }
  return res;
}
long long int modi(long long int a) {
  long long int m = MODA, s = 1, p = 0;
  while (a > 1) {
    long long int q = a / m, t = m;
    m = a % m;
    a = t;
    t = p;
    p = s - q * p;
    s = t;
  }
  return s >= 0 ? s : s + MODA;
}
void generatePrimeFactors(long long int n) {
  primefactors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primefactors.push_back(i);
      while (n % i == 0) n = n / i;
    }
  }
  if (n != 1) primefactors.push_back(n);
}
long long int ncr(long long int n, long long int r) {
  if (r == 0) return 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % MODA;
  return (fact[n] * modi(fact[r]) % MODA * modi(fact[n - r]) % MODA) % MODA;
}
void generateFactors(long long int n) {
  factors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      if (n / i != i) factors.push_back(n / i);
    }
  }
  factors.push_back(1);
  factors.push_back(n);
  sort(factors.begin(), factors.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i, n, k, flag = 0;
  cin >> n >> k;
  string s;
  cin >> s;
  long long int d = (n / k) * k;
  string check = s.substr(0, k), subwindow = "";
  for (i = d; i < n; i += 1) subwindow += s[i];
  int len = subwindow.length();
  for (i = 1; i < n / k; i++) {
    if (s.substr(i * k, k) == check) {
      continue;
    }
    if (s.substr(i * k, k) < check) {
      flag = -1;
      break;
    }
    if (s.substr(i * k, k) > check) {
      flag = 1;
      break;
    }
  }
  if (check.substr(0, len) < subwindow && flag != -1) flag = 1;
  if (flag == 1) {
    if (check[k - 1] == '9') {
      int carry = 1;
      for (i = k - 1; i >= 0; i--) {
        if (carry == 0) break;
        int num = carry + (check[i] - '0');
        if (num > 9) {
          carry = 1;
          num = num % 10;
        } else
          carry = 0;
        check[i] = (char)(num + '0');
      }
      if (carry) check = (char)(carry + '0') + check;
    } else
      check[k - 1] = (char)((int)check[k - 1] + 1);
  }
  cout << n << "\n";
  for (int i = 0; i < n; i++) cout << check[i % k];
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const long long int MODA = 1000000007;
vector<long long int> primefactors;
vector<long long int> factors;
long long int fact[1025] = {};
bool isPowerTwo(long long int x) { return (x && !(x & (x - 1))); }
long long int modmul(long long int a, long long int b) {
  return ((a % MODA) * (b % MODA)) % MODA;
}
long long int modadd(long long int a, long long int b) {
  return ((a % MODA) + (b % MODA) + MODA) % MODA;
}
long long int modsub(long long int a, long long int b) {
  return ((a % MODA) - (b % MODA) + MODA) % MODA;
}
bool isSubstring(string s1, string s2) {
  if (s1.find(s2) != string::npos)
    return true;
  else
    return false;
}
void generateFactorial(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) fact[i] = (i * 1ll * fact[i - 1]);
}
void generateFactorialMOD(long long int n) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++)
    fact[i] = (i * 1ll * fact[i - 1]) % MODA;
}
bool isPrime(long long int n) {
  if (n < 2) return false;
  for (long long int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
long long int powermod(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MODA;
    a = (a * a) % MODA;
    b >>= 1;
  }
  return res;
}
long long int modi(long long int a) {
  long long int m = MODA, s = 1, p = 0;
  while (a > 1) {
    long long int q = a / m, t = m;
    m = a % m;
    a = t;
    t = p;
    p = s - q * p;
    s = t;
  }
  return s >= 0 ? s : s + MODA;
}
void generatePrimeFactors(long long int n) {
  primefactors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      primefactors.push_back(i);
      while (n % i == 0) n = n / i;
    }
  }
  if (n != 1) primefactors.push_back(n);
}
long long int ncr(long long int n, long long int r) {
  if (r == 0) return 1;
  fact[0] = 1;
  for (int i = 1; i <= n; i++) fact[i] = fact[i - 1] * i % MODA;
  return (fact[n] * modi(fact[r]) % MODA * modi(fact[n - r]) % MODA) % MODA;
}
void generateFactors(long long int n) {
  factors.clear();
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      if (n / i != i) factors.push_back(n / i);
    }
  }
  factors.push_back(1);
  factors.push_back(n);
  sort(factors.begin(), factors.end());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i, n, k, flag = 0;
  cin >> n >> k;
  string s;
  cin >> s;
  long long int d = (n / k) * k;
  string check = s.substr(0, k), subwindow = "";
  for (i = d; i < n; i += 1) subwindow += s[i];
  int len = subwindow.length();
  for (i = 1; i < n / k; i++) {
    if (s.substr(i * k, k) == check) {
      continue;
    }
    if (s.substr(i * k, k) < check) {
      flag = -1;
      break;
    }
    if (s.substr(i * k, k) > check) {
      flag = 1;
      break;
    }
  }
  if (check.substr(0, len) < subwindow && flag != -1) flag = 1;
  if (flag == 1) {
    if (check[k - 1] == '9') {
      int carry = 1;
      for (i = k - 1; i >= 0; i--) {
        if (carry == 0) break;
        int num = carry + (check[i] - '0');
        if (num > 9) {
          carry = 1;
          num = num % 10;
        } else
          carry = 0;
        check[i] = (char)(num + '0');
      }
      if (carry) check = (char)(carry + '0') + check;
    } else
      check[k - 1] = (char)((int)check[k - 1] + 1);
  }
  cout << n << "\n";
  for (int i = 0; i < n; i++) cout << check[i % k];
  return 0;
}

```