#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-4;
const double pi = acos(-1.0);
void debug() { cerr << '\n'; }
template <typename T, typename... Ts>
void debug(T x, Ts... y) {
  cerr << "\033[31m" << x << "\033[0m";
  if (sizeof...(y) > 0) cerr << ' ';
  debug(y...);
}
template <typename T>
void debug(const T &a, int l, int r, char c) {
  for (int i = l; i <= r; ++i)
    cerr << "\033[31m" << a[i] << "\033[0m" << (i == r ? '\n' : c);
}
template <typename T>
void debug(vector<T> a) {
  for (int i = 0; i < (int)a.size(); ++i)
    cerr << "\033[31m" << a[i] << "\033[31m"
         << " \n"[i == ((int)a.size() - 1)];
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long qpow(long long x, long long y) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans *= x;
    x *= x;
  }
  return ans;
}
long long qpow(long long x, long long y, int MOD) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans = ans * x % MOD;
    x = x * x % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long X = x;
  x = y;
  y = X - a / b * y;
}
long long inv(long long x, int mod = MOD) { return qpow(x, mod - 2, mod); }
long long m_m(long long a, int mod = MOD) { return (a % mod + mod) % mod; }
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  string t = s.substr(0, k);
  for (int i = (k); i < (n); ++i) t += t[i - k];
  cout << n << '\n';
  if (t >= s) return cout << t, 0;
  for (int i = (k - 1); i > (-1); --i) {
    if (t[i] == '9')
      t[i] = '0';
    else {
      ++t[i];
      break;
    }
  }
  for (int i = (k); i < (n); ++i) t[i] = t[i - k];
  cout << t;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#pragma GCC optimize(2)
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-4;
const double pi = acos(-1.0);
void debug() { cerr << '\n'; }
template <typename T, typename... Ts>
void debug(T x, Ts... y) {
  cerr << "\033[31m" << x << "\033[0m";
  if (sizeof...(y) > 0) cerr << ' ';
  debug(y...);
}
template <typename T>
void debug(const T &a, int l, int r, char c) {
  for (int i = l; i <= r; ++i)
    cerr << "\033[31m" << a[i] << "\033[0m" << (i == r ? '\n' : c);
}
template <typename T>
void debug(vector<T> a) {
  for (int i = 0; i < (int)a.size(); ++i)
    cerr << "\033[31m" << a[i] << "\033[31m"
         << " \n"[i == ((int)a.size() - 1)];
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long qpow(long long x, long long y) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans *= x;
    x *= x;
  }
  return ans;
}
long long qpow(long long x, long long y, int MOD) {
  long long ans = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) ans = ans * x % MOD;
    x = x * x % MOD;
  }
  return ans;
}
void exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  exgcd(b, a % b, x, y);
  long long X = x;
  x = y;
  y = X - a / b * y;
}
long long inv(long long x, int mod = MOD) { return qpow(x, mod - 2, mod); }
long long m_m(long long a, int mod = MOD) { return (a % mod + mod) % mod; }
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  string t = s.substr(0, k);
  for (int i = (k); i < (n); ++i) t += t[i - k];
  cout << n << '\n';
  if (t >= s) return cout << t, 0;
  for (int i = (k - 1); i > (-1); --i) {
    if (t[i] == '9')
      t[i] = '0';
    else {
      ++t[i];
      break;
    }
  }
  for (int i = (k); i < (n); ++i) t[i] = t[i - k];
  cout << t;
  return 0;
}

```