Problem:
Score: 100 points

Problem Statement

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.

Integers M_1, D_1, M_2, and D_2 will be given as input.

It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.

Determine whether the date 2019-M_1-D_1 is the last day of a month.

Constraints

Both 2019-M_1-D_1 and 2019-M_2-D_2 are valid dates in the Gregorian calendar.

The date 2019-M_2-D_2 follows 2019-M_1-D_1.

Input

Input is given from Standard Input in the following format:

M_1 D_1
M_2 D_2

Output

If the date 2019-M_1-D_1 is the last day of a month, print 1; otherwise, print 0.

Sample Input 1

11 16
11 17

Sample Output 1

0

November 16 is not the last day of a month.

Sample Input 2

11 30
12 1

Sample Output 2

1

November 30 is the last day of November.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <array>
#include <set>
#include <map>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cstdint>
#include <cassert>
#include <random>

using namespace std;
using i64 = int64_t;
using i32 = int32_t;
template<class T, class U> void init_n(vector<T>& v, size_t n, U x) 
{ v = vector<T>(n, x); }
template<class T> void init_n(vector<T>& v, size_t n) { init_n(v, n, T()); }
template<class T> void read_n(vector<T>& v, size_t n, size_t o = 0) 
{ v = vector<T>(n+o); for (size_t i=o; i<n+o; ++i) cin >> v[i]; }
template<class T> void read_n(T a[], size_t n, size_t o = 0)
{ for (size_t i=o; i<n+o; ++i) cin >> a[i]; }
template<class T> T gabs(const T& x) { return max(x, -x); }
#define abs gabs

int m1, d1, m2, d2;
int main() {
  cin >> m1 >> d1 >> m2 >> d2;

  cout << (m1 == m2 ? '0' : '1') << endl;
  return 0;
}