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 sc=new Scanner(System.in);
int N= sc.nextInt();
int T= sc.nextInt();
int c[]=new int[N];
int t[]=new int[N];
for(int n=0;n<N;n++) {
c[n]=sc.nextInt();
t[n]=sc.nextInt();
 }
int cost=0;
boolean home=false;
for(int k=0;k<N;k++) {
	if(!home) {
	if(t[k]<=T) {
	cost=c[k];
	home=true;
	  }
	 }
 if(c[k]<cost) {
		if(t[k]<=T){
			cost=c[k];
		}
	}
}
System.out.println((home)?cost:"TLE");
 }
  }