#include <bits/stdc++.h>
using namespace std;
mt19937 rng(static_cast<unsigned int>(
    chrono::steady_clock::now().time_since_epoch().count()));
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(const num_t& lhs, const num_t& rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t lcm(const num_t& lhs, const num_t& rhs) {
  return lhs / ::gcd(lhs, rhs) * rhs;
}
template <typename num_t>
num_t pw(num_t n, num_t k, const num_t& mod) {
  if (k == -1) k = mod - 2;
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res % mod;
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const long long ll_inf = (long long)inf * inf * 9 + 7;
const int maxn = 500000 + 7;
void solve() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  string temp = s.substr(0, k);
  for (int i = k; i < n; ++i) temp += temp[i - k];
  cout << n << '\n';
  if (temp >= s) {
    cout << temp << '\n';
    return;
  }
  temp = s.substr(0, k);
  int mem = 1;
  for (int i = k - 1; i >= 0; --i) {
    int cur = s[i] - '0' + mem;
    temp[i] = (cur % 10) + '0';
    mem = cur / 10;
  }
  for (int i = k; i < n; ++i) temp += temp[i - k];
  cout << temp << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

### 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;
mt19937 rng(static_cast<unsigned int>(
    chrono::steady_clock::now().time_since_epoch().count()));
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(const num_t& lhs, const num_t& rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t lcm(const num_t& lhs, const num_t& rhs) {
  return lhs / ::gcd(lhs, rhs) * rhs;
}
template <typename num_t>
num_t pw(num_t n, num_t k, const num_t& mod) {
  if (k == -1) k = mod - 2;
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res % mod;
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const long long ll_inf = (long long)inf * inf * 9 + 7;
const int maxn = 500000 + 7;
void solve() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  string temp = s.substr(0, k);
  for (int i = k; i < n; ++i) temp += temp[i - k];
  cout << n << '\n';
  if (temp >= s) {
    cout << temp << '\n';
    return;
  }
  temp = s.substr(0, k);
  int mem = 1;
  for (int i = k - 1; i >= 0; --i) {
    int cur = s[i] - '0' + mem;
    temp[i] = (cur % 10) + '0';
    mem = cur / 10;
  }
  for (int i = k; i < n; ++i) temp += temp[i - k];
  cout << temp << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}

```