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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int T = sc.nextInt();
		List<Integer> costs = new ArrayList<>();
		List<Integer> times = new ArrayList<>();
		List<Integer> tmpCosts = new ArrayList<>();
		for(int a = 0 ; a < N ; a++){
			costs.add(sc.nextInt());
			times.add(sc.nextInt());
		}
		sc.close();
		for(int b = 0 ; b < N ; b++){
			if(times.get(b) <= T){
				tmpCosts.add(costs.get(b));
			}
		}
		if(tmpCosts.isEmpty()){
			System.out.println("TLE");
		}else{		
			Collections.sort(tmpCosts);
			System.out.println(tmpCosts.get(0));
		}
	 }
}