Problem: Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records. 

The records are expressed as a string s of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer p is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer p is considered a period of string s, if for all 1 ≤ i ≤ \lvert s \rvert - p, the i-th and (i + p)-th characters of s are the same. Here \lvert s \rvert is the length of s.

Input

The first line contains two space-separated integers n and p (1 ≤ p ≤ n ≤ 2000) — the length of the given string and the supposed period, respectively.

The second line contains a string s of n characters — Mino's records. s only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that p is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

Input

10 7
1.0.1.0.1.


Output

1000100010


Input

10 6
1.0.1.1000


Output

1001101000


Input

10 9
1........1


Output

No

Note

In the first example, 7 is not a period of the resulting string because the 1-st and 8-th characters of it are different.

In the second example, 6 is not a period of the resulting string because the 4-th and 10-th characters of it are different.

In the third example, 9 is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e15;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;
const double PI = acos(-1.0);
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int N = 2005;
long long n, p;
bool have[N][3];
string s;
bool solve() {
  bool ok = false;
  for (int i = 0; i < min(n - p, p); ++i) {
    if (have[i][0] && have[i][1]) {
      ok = true;
      for (int j = i; j < n; j += p)
        if (s[j] == '.') s[j] = '0';
    } else if (have[i][0]) {
      if (have[i][2]) {
        ok = true;
        for (int j = i; j < n; j += p) {
          if (s[j] == '.') s[j] = '1';
        }
      }
    } else if (have[i][1]) {
      if (have[i][2]) {
        ok = true;
        for (int j = i; j < n; j += p) {
          if (s[j] == '.') s[j] = '0';
        }
      }
    } else {
      s[i] = '1';
      for (int j = i + p; j < n; j += p) s[j] = '0';
      ok = true;
    }
  }
  for (int i = 0; i < n; ++i)
    if (s[i] == '.') s[i] = '0';
  return ok;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  cin >> n >> p >> s;
  for (int i = 0; i < n; ++i)
    if (s[i] != '.')
      have[i % p][s[i] - '0'] = true;
    else
      have[i % p][2] = true;
  if (solve())
    cout << s << "\n";
  else
    cout << "No"
         << "\n";
}