Jamie loves sleeping. One day, he decides that he needs to wake up at exactly hh: mm. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every x minutes until hh: mm is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.

A time is considered lucky if it contains a digit '7'. For example, 13: 07 and 17: 27 are lucky, while 00: 48 and 21: 34 are not lucky.

Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a lucky time Jamie can set so that he can wake at hh: mm.

Formally, find the smallest possible non-negative integer y such that the time representation of the time x·y minutes before hh: mm contains the digit '7'.

Jamie uses 24-hours clock, so after 23: 59 comes 00: 00.

Input

The first line contains a single integer x (1 ≤ x ≤ 60).

The second line contains two two-digit integers, hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59).

Output

Print the minimum number of times he needs to press the button.

Examples

Input

3
11 23


Output

2


Input

5
01 07


Output

0

Note

In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.

In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int x, hh, mm;
  char s1[5], s2[5];
  scanf("%d", &x);
  scanf("%s %s", s1, s2);
  int h1, h2, m1, m2;
  h1 = s1[0] - '0';
  h2 = s1[1] - '0';
  m1 = s2[0] - '0';
  m2 = s2[1] - '0';
  if (h2 == 7 || m2 == 7)
    printf("0\n");
  else {
    int t = 0;
    while (m2 != 7 && h2 != 7) {
      t++;
      if (x > m2) {
        if (x > (m1 * 10 + m2)) {
          if (h2 >= 1) {
            h2 -= 1;
            int xx = 60 + m1 * 10 + m2 - x;
            m1 = xx / 10;
            m2 = xx % 10;
          } else {
            if (h1 >= 1) {
              h1 -= 1;
              h2 = 9;
              int xx = 60 + m1 * 10 + m2 - x;
              m1 = xx / 10;
              m2 = xx % 10;
            } else {
              h1 = 2;
              h2 = 3;
              int xx = 60 + m1 * 10 + m2 - x;
              m1 = xx / 10;
              m2 = xx % 10;
            }
          }
        } else {
          int x1 = x / 10;
          int x2 = x % 10;
          if (x2 > m2) {
            m2 = 10 + m2 - x2;
            m1 = m1 - 1 - x1;
          } else {
            m2 -= x2;
            m1 -= x1;
          }
        }
      } else
        m2 -= x;
    }
    printf("%d\n", t);
  }
  return 0;
}
