Problem:
Score : 200 points

Problem Statement

When Mr. X is away from home, he has decided to use his smartwatch to search the best route to go back home, to participate in ABC.

You, the smartwatch, has found N routes to his home.

If Mr. X uses the i-th of these routes, he will get home in time t_i at cost c_i.

Find the smallest cost of a route that takes not longer than time T.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq T \leq 1000

1 \leq c_i \leq 1000

1 \leq t_i \leq 1000

The pairs (c_i, t_i) are distinct.

Input

Input is given from Standard Input in the following format:

N T
c_1 t_1
c_2 t_2
:
c_N t_N

Output

Print the smallest cost of a route that takes not longer than time T.

If there is no route that takes not longer than time T, print TLE instead.

Sample Input 1

3 70
7 60
1 80
4 50

Sample Output 1

4

The first route gets him home at cost 7.

The second route takes longer than time T = 70.

The third route gets him home at cost 4.

Thus, the cost 4 of the third route is the minimum.

Sample Input 2

4 3
1 1000
2 4
3 1000
4 500

Sample Output 2

TLE

There is no route that takes not longer than time T = 3.

Sample Input 3

5 9
25 8
5 9
4 10
1000 1000
6 1

Sample Output 3

5

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

#include <bits/stdc++.h>
#define ll long long int
#define MOD 1000000007
#define INF 1e18
#define PI 3.14159265358979

using namespace std;

ll gcd(ll a, ll b);
ll power(ll n, ll k);
ll inv(ll n);
ll genfact(ll n, ll k);
ll fact(ll n);
ll nck(ll n, ll k);

int main(void){
  int n, t;
  cin >> n >> t;
  int m = 1e9;
  int a, b;
  for (int i = 0; i < n; i++){
    cin >> a >> b;
    if (b <= t)
      m = min(m, a);
  }

  cout << m << endl;

  return 0;
}

ll gcd(ll a, ll b){
  if (a < b)
    return gcd(b, a);
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}

ll power(ll n, ll k){
  ll res = 1;
  while(k > 0){
    if (k & 1) 
      res = res * n % MOD;
    n = n * n % MOD;
    k >>= 1;
  }
  return res;
}

ll genfact(ll n, ll k){

  ll res = 1;
  while (k >= 0){
    res = (res * (n - k)) % MOD;
    k--;
  }

  return res;
}

ll inv(ll n){
  return power(n, MOD - 2);
}

ll fact(ll n){
  return genfact(n, n - 1);
}

ll nck(ll n, ll k){
  if (k == 0)
    return 1;
  if (n <= 0)
    return 0;

  return ((genfact(n, k) % MOD) * inv(fact(k))) % MOD;
}