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.Scanner;

class Main{
	public static void main(String[] args) {



		Scanner scanner = new Scanner(System.in);
		int N = scanner.nextInt();
		int T = scanner.nextInt();
		int[] c = new int[N] ;
		int[] t = new int[N] ;


		for(int i = 0;i < N;i++){
			c[i] =  scanner.nextInt();
			t[i] =  scanner.nextInt();

		}
		int ans = 100000;
		for(int i= 0;i < N;i++){
			if(t[i] <= T){
			ans = 	Math.min(ans, c[i]);
			}
		}
		if(ans == 100000){
			System.out.println("TLE");
		}else{
		System.out.println(ans);


		}
	}


}

class Pair implements Comparable{
	int from;
	int end;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}