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;
const unsigned long long prime = 153;
const double eps = 1e-9;
const double pi = acos(-1);
const long long mod = 1e9 + 7;
int px[] = {0, 0, 1, -1, 1, -1, 1, -1};
int py[] = {1, -1, 0, 0, 1, 1, -1, -1};
int kx[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky[] = {2, -2, 1, -1, 2, -2, 1, -1};
bool check(int b, int c) {
  while (b) {
    if (b % 10 == 7) return 1;
    b /= 10;
  }
  while (c) {
    if (c % 10 == 7) return 1;
    c /= 10;
  }
  return 0;
}
int main() {
  int test, t = 1;
  int a, b, c, d, v, angle, time, y, e, f, g, h;
  double first, second, third;
  double one, two, three, four, five, six, seven, eight, nine, ten;
  scanf("%d", &a);
  scanf("%d%d", &b, &c);
  bool r = 0;
  int press(0);
  do {
    bool de = check(b, c);
    if (de) break;
    d = b * 60 + c;
    d -= a;
    c = d % 60;
    b = d / 60;
    press++;
    if (d < 0) {
      v = 24 * 60 + d;
      c = v % 60;
      b = v / 60;
    }
  } while (1);
  printf("%d", press);
}
