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 = Integer.parseInt(sc.next());
	int T = Integer.parseInt(sc.next());
	int c[] = new int[N];
	int t[] = new int[N];
	for (int i = 0; i < N; i++) {
	    c[i] = Integer.parseInt(sc.next());
	    t[i] = Integer.parseInt(sc.next());
	}

	int minT = t[0];
	for (int i = 1; i< N; i++) {
	    int v = t[i];
	    if (v < minT) {
		minT = v;
	    }
	}

	int minC = c[0];
	for (int i = 0; i < N; i++) {
	    int v = t[i];
	    int w = c[i];
	    if (v <= T && w < minC) {
		minC = w;
	    }
	}

	if (T < minT) {
	    System.out.println("TLE");
	} else {
	    System.out.println(minC);
	}
    }
}