In problems on strings one often has to find a string with some particular properties. The problem authors were reluctant to waste time on thinking of a name for some string so they called it good. A string is good if it doesn't have palindrome substrings longer than or equal to d. 

You are given string s, consisting only of lowercase English letters. Find a good string t with length |s|, consisting of lowercase English letters, which is lexicographically larger than s. Of all such strings string t must be lexicographically minimum.

We will call a non-empty string s[a ... b] = sasa + 1... sb (1 ≤ a ≤ b ≤ |s|) a substring of string s = s1s2... s|s|.

A non-empty string s = s1s2... sn is called a palindrome if for all i from 1 to n the following fulfills: si = sn - i + 1. In other words, palindrome read the same in both directions.

String x = x1x2... x|x| is lexicographically larger than string y = y1y2... y|y|, if either |x| > |y| and x1 = y1, x2 = y2, ... , x|y| = y|y|, or there exists such number r (r < |x|, r < |y|), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 > yr + 1. Characters in such strings are compared like their ASCII codes.

Input

The first line contains integer d (1 ≤ d ≤ |s|).

The second line contains a non-empty string s, its length is no more than 4·105 characters. The string consists of lowercase English letters.

Output

Print the good string that lexicographically follows s, has the same length and consists of only lowercase English letters. If such string does not exist, print "Impossible" (without the quotes).

Examples

Input

3
aaaaaaa


Output

aabbcaa


Input

3
zzyzzzz


Output

Impossible


Input

4
abbabbbabbb


Output

abbbcaaabab
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int exp(int x, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * x % mod;
    x = 1LL * x * x % mod;
    n >>= 1;
  }
  return ret;
}
int inv(int x) { return exp(x, mod - 2); }
int po[400010], invp[400010];
int N, D;
string S, T;
int sum1[400010], sum2[400010], offset;
int calc1(int l, int r) {
  return (sum1[r] + mod -
          1LL * (l == 0 ? 0 : sum1[l - 1]) * po[r - l + 1] % mod) %
         mod;
}
int calc2(int l, int r) {
  return (1LL * (sum2[l] + offset) * invp[l] % mod + mod -
          (1LL * (sum2[r + 1] + offset) * invp[r + 1] % mod) * po[r - l + 1] %
              mod) %
         mod;
}
bool chk(int x) {
  if (x + 1 >= D && calc1(x - D + 1, x) == calc2(x - D + 1, x)) return true;
  if (x + 1 >= D + 1 && calc1(x - D, x) == calc2(x - D, x)) return true;
  return false;
}
bool f(int x) {
  int t = 1;
  for (int i = x; i < N; i++) {
    bool ok = false;
    for (int j = t ? S[i] - 'a' + 1 : 0; j < 26; j++) {
      T.push_back('a' + j);
      sum1[i] = 1LL * (i == 0 ? 0 : sum1[i - 1]) * 27 % mod + T[i] - 'a' + 1;
      sum1[i] %= mod;
      offset += 1LL * po[i] * (T[i] - 'a' + 1) % mod;
      offset %= mod;
      sum2[i + 1] = mod - offset;
      sum2[i + 1] %= mod;
      if (chk(i)) {
        offset += mod - 1LL * po[i] * (T[i] - 'a' + 1) % mod;
        offset %= mod;
        T.pop_back();
        continue;
      } else {
        ok = true;
        t = 0;
        break;
      }
    }
    if (!ok) return false;
    if (i == N - 1 && !t) return true;
  }
  return false;
}
int main() {
  po[0] = 1;
  for (int i = 1; i < 400010; i++) {
    po[i] = 1LL * po[i - 1] * 27 % mod;
  }
  for (int i = 0; i < 400010; i++) {
    invp[i] = inv(po[i]);
  }
  cin >> D >> S;
  N = S.size();
  T.clear();
  int p = N;
  for (int i = 0; i < N; i++) {
    T.push_back(S[i]);
    sum1[i] = 1LL * (i == 0 ? 0 : sum1[i - 1]) * 27 % mod + T[i] - 'a' + 1;
    sum1[i] %= mod;
    offset += 1LL * po[i] * (T[i] - 'a' + 1) % mod;
    offset %= mod;
    sum2[i + 1] = mod - offset;
    sum2[i + 1] %= mod;
    if (chk(i)) {
      p = i;
      offset += mod - 1LL * po[i] * (T[i] - 'a' + 1) % mod;
      offset %= mod;
      T.pop_back();
      break;
    }
  }
  for (int i = p; i >= 0; i--) {
    if (f(i)) {
      cout << T;
      return 0;
    }
    if (i == 0) {
      printf("Impossible");
      return 0;
    }
    offset += mod - 1LL * po[i - 1] * (T[i - 1] - 'a' + 1) % mod;
    offset %= mod;
    T.pop_back();
  }
}
