#include <bits/stdc++.h>
using namespace std;
using namespace std::chrono;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int mod = 1e9 + 7;
long long binpow(long long num, long long pow);
int extendedEuclid(int a, int b, int &x, int &y);
inline int mul(int a, int b) { return (a * 1ll * b) % mod; }
inline int add(int a, int b) { return (a + 0ll + b) % mod; }
inline int sub(int a, int b) { return ((a + 0ll - b) % mod + mod) % mod; }
inline void intarr(vector<int> &a) {
  for (auto &i : a) cin >> i;
}
inline void larr(vector<long long> &a) {
  for (auto &i : a) cin >> i;
}
inline void solve() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  int ok = 1;
  for (int i = (int)0; i < (int)n; i++) {
    if (s[i] != s[i % k]) {
      if (s[i] > s[i % k]) ok = 0;
      break;
    }
  }
  if (ok) {
    cout << n << endl;
    for (int i = (int)0; i < (int)n; i++) cout << s[i % k];
    cout << endl;
    return;
  }
  int pos = k - 1;
  s[pos]++;
  while (s[pos] == '9' + 1) {
    s[pos - 1]++;
    s[pos] = '0';
    pos--;
  }
  cout << n << endl;
  for (int i = (int)0; i < (int)n; i++) cout << s[i % k];
  cout << endl;
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) solve();
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  cerr << duration.count() << " microseconds\n";
  return 0;
}
long long binpow(long long num, long long pow) {
  if (pow == 0) return 1;
  long long val = binpow(num, pow / 2);
  val *= val;
  if (pow & 1) val *= num;
  return val;
}
int extendedEuclid(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  int x1, y1;
  int d = extendedEuclid(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
using namespace std::chrono;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int mod = 1e9 + 7;
long long binpow(long long num, long long pow);
int extendedEuclid(int a, int b, int &x, int &y);
inline int mul(int a, int b) { return (a * 1ll * b) % mod; }
inline int add(int a, int b) { return (a + 0ll + b) % mod; }
inline int sub(int a, int b) { return ((a + 0ll - b) % mod + mod) % mod; }
inline void intarr(vector<int> &a) {
  for (auto &i : a) cin >> i;
}
inline void larr(vector<long long> &a) {
  for (auto &i : a) cin >> i;
}
inline void solve() {
  int n, k;
  string s;
  cin >> n >> k >> s;
  int ok = 1;
  for (int i = (int)0; i < (int)n; i++) {
    if (s[i] != s[i % k]) {
      if (s[i] > s[i % k]) ok = 0;
      break;
    }
  }
  if (ok) {
    cout << n << endl;
    for (int i = (int)0; i < (int)n; i++) cout << s[i % k];
    cout << endl;
    return;
  }
  int pos = k - 1;
  s[pos]++;
  while (s[pos] == '9' + 1) {
    s[pos - 1]++;
    s[pos] = '0';
    pos--;
  }
  cout << n << endl;
  for (int i = (int)0; i < (int)n; i++) cout << s[i % k];
  cout << endl;
}
int main() {
  auto start = high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) solve();
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  cerr << duration.count() << " microseconds\n";
  return 0;
}
long long binpow(long long num, long long pow) {
  if (pow == 0) return 1;
  long long val = binpow(num, pow / 2);
  val *= val;
  if (pow & 1) val *= num;
  return val;
}
int extendedEuclid(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1, y = 0;
    return a;
  }
  int x1, y1;
  int d = extendedEuclid(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}

```