Problem:
Score : 200 points

Problem Statement

You are given two positive integers A and B. Compare the magnitudes of these numbers.

Constraints

1 ≤ A, B ≤ 10^{100}

Neither A nor B begins with a 0.

Input

Input is given from Standard Input in the following format:

A
B

Output

Print GREATER if A>B, LESS if A<B and EQUAL if A=B.

Sample Input 1

36
24

Sample Output 1

GREATER

Since 36>24, print GREATER.

Sample Input 2

850
3777

Sample Output 2

LESS

Sample Input 3

9720246
22516266

Sample Output 3

LESS

Sample Input 4

123456789012345678901234567890
234567890123456789012345678901

Sample Output 4

LESS

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

#include <iostream>      // cout, endl, cin
#include <string>        // string, to_string, stoi
#include <vector>        // vector
#include <algorithm>     // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility>       // pair, make_pair
#include <tuple>         // tuple, make_tuple
#include <cstdint>       // int64_t, int*_t
#include <cstdio>        // printf
#include <map>           // map
#include <cmath>         // sqrt
#include <queue>         // queue, priority_queue
#include <set>           // set
#include <stack>         // stack
#include <deque>         // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset>        // bitset
#include <bits/stdc++.h> // isupper, islower, isdigit, toupper, tolower
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rep2(i, a, n) for (int i = (a); i < (n); i++)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> Pl;
typedef pair<int, int> P;
typedef tuple<ll, ll, string> Tlls;

// Greatest Common Divisor (最大公約数)
ll gcd(ll a, ll b)
{
  // Eucledean algorithm
  while (a)
  {
    if (a < b)
    {
      std::swap(a, b);
    }
    a %= b;
  }
  return b;
}

// Least Common Multiple 最小公倍数
ll lcm(ll a, ll b)
{
  return a / gcd(a, b) * b;
}

// key: value of prime factor素因数分解
// value: number of prime factor
map<ll, ll> prime_factorization(ll n)
{
  map<ll, ll> prime_factors;
  for (ll i = 2; i * i <= n; i++)
  {
    while (n % i == 0)
    {
      prime_factors[i]++;
      n /= i;
    }
  }
  if (n != 1)
    prime_factors[n] = 1;
  return prime_factors;
}

// 約数リスト約数の列挙
vector<ll> get_divisors(const ll n)
{
  vector<ll> divs;
  for (ll i = 1; i < (ll)sqrt((ld)n) + 1; i++)
  {
    if (n % i == 0)
    {
      divs.push_back(i);
      if (i != n / i)
      {
        divs.push_back(n / i);
      }
    }
  }

  // 必要に応じてソート
  // sort(divs.begin(), divs.end());

  return divs;
}

int main()
{

  string A, B;
  cin >> A >> B;
  if (A.length() > B.length())
  {
    cout << "GREATER" << endl;
    return 0;
  }
  else if (A.length() < B.length())
  {
    cout << "LESS" << endl;
    return 0;
  }
  rep(i, A.length())
  {
    if (A.at(i) > B.at(i))
    {
      cout << "GREATER" << endl;
      return 0;
    }
    else if (A.at(i) < B.at(i))
    {
      cout << "LESS" << endl;
      return 0;
    }
  }
  cout << "EQUAL" << endl;

  return 0;
}