Problem: You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.

You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.

For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.

Input

The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.

The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.

Output

The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.

Examples

Input

24
17:30


Output

17:30


Input

12
17:30


Output

07:30


Input

24
99:99


Output

09:09

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const long long LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const long long mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101",
                     "0110", "0111", "1000", "1001", "1010", "1011",
                     "1100", "1101", "1110", "1111"};
inline long long gcd(long long a, long long b) {
  return b == 0 ? a : gcd(b, a % b);
}
inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b) { return a < b ? a : b; }
inline int Max(int a, int b) { return a > b ? a : b; }
inline long long Min(long long a, long long b) { return a < b ? a : b; }
inline long long Max(long long a, long long b) { return a > b ? a : b; }
inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; }
int main() {
  while (scanf("%d", &n) == 1) {
    int h, s;
    scanf("%d:%d", &h, &s);
    if (n == 12) {
      if (h > 12) {
        int t = h % 10;
        int tt = h / 10 % 10;
        if (t == 0)
          h = 10;
        else if (t < 3)
          h = 10 + t;
        else if (tt < 2)
          h = tt * 10 + 0;
        else
          h = t;
      } else if (h == 0)
        h = 10;
    } else {
      if (h > 23) {
        int t = h % 10;
        int tt = h / 10 % 10;
        if (t == 0)
          h = 10;
        else if (t < 4)
          h = 10 + t;
        else if (tt < 3)
          h = tt * 10 + 0;
        else
          h = t;
      }
    }
    int t = s % 10;
    if (s >= 60) s = 10 + t;
    printf("%02d:%02d\n", h, s);
  }
  return 0;
}