Problem:
Score : 100 points

Problem Statement

Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.

The problem has N test cases, all of which must be passed to get an AC verdict.

Takahashi's submission has passed M cases out of the N test cases.

Determine whether Takahashi's submission gets an AC.

Constraints

1 \leq N \leq 100

0 \leq M \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

If Takahashi's submission gets an AC, print Yes; otherwise, print No.

Sample Input 1

3 3

Sample Output 1

Yes

All three test cases have been passed, so his submission gets an AC.

Sample Input 2

3 2

Sample Output 2

No

Only two out of the three test cases have been passed, so his submission does not get an AC.

Sample Input 3

1 1

Sample Output 3

Yes

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

#include <algorithm>
#include <bitset>
#include <cmath>
#include <complex>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <vector>

const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;

using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;

typedef pair<ll, ll> pll;
bool paircomp(const pll &a, const pll &b) {
  if (a.first == b.first)
    return a.second < b.second;
  return a.first < b.first;
}

#define REP(i, n) for (ll i = 0; i < (n); i++)
#define RREP(i, n) for (ll i = (n)-1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define AUTO(i, m) for (auto &i : m)
#define ALL(a) (a).begin(), (a).end()
#define MAX(vec) *std::max_element(vec.begin(), vec.end())
#define MIN(vec) *std::min_element(vec.begin(), vec.end())
#define ARGMAX(vec)                                                            \
  std::distance(vec.begin(), std::max_element(vec.begin(), vec.end()))
#define ARGMIN(vec)                                                            \
  std::distance(vec.begin(), std::min_element(vec.begin(), vec.end()))
#define REV(T) greater<T>()
#define PQ(T) priority_queue<T, vector<T>, greater<T>>
#define VVL(a, b, c) vector<vector<ll>>(a, vector<ll>(b, c))
#define VV(T, a, b, c) vector<vector<T>>(a, vector<T>(b, c))
#define VVVL(a, b, c, d)                                                       \
  vector<vector<vector<ll>>>(a, vector<vector<ll>>(b, vector<ll>(c, d)))
#define VVV(T, a, b, c, d)                                                     \
  vector<vector<vector<T>>>(a, vector<vector<T>>(b, vector<T>(c, d)))
#define SP(a) setprecision(a)
#define SQRT(a) sqrt((long double)(a))
#define DPOW(a, b) pow((long double)(a), (long double)(b))
#define UNIQUE(vec)                                                            \
  do {                                                                         \
    sort(ALL((vec)));                                                          \
    (vec).erase(std::unique(ALL((vec))), (vec).end());                         \
  } while (0)

ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}

int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll N, M;
  cin >> N >> M;
  vl A;
  if (N == M)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}