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();
    int c[] = new int[N];
    int t[] = new int[N];
    int cmin = 1001;
    int twithin = 1001;
    
    for(int i= 0;i<N;i++){
      c[i]=sc.nextInt();
      t[i]=sc.nextInt();
      
      if(t[i]<=T && cmin>c[i]){
        //tがT以内かつcが最小になるまで値をサーチ
        twithin=t[i];
        cmin=c[i];
      }
        
    }
    if(twithin==1001){
      System.out.println("TLE");
    }else{
      System.out.println(cmin);
    }
  }
}