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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
typedef struct return_home{
	int cost;
	int time;
}Ret;

int main(void) {
	int n, t;
	Ret ret[100];
	Ret ans = { 0, 0 };

	scanf("%d %d", &n, &t);
	for (int i = 0; i < n; i++) {
		scanf("%d %d", &ret[i].cost, &ret[i].time);
		if (t >= ret[i].time) {
			if (ans.cost == 0) {
				ans = ret[i];
			}
			else {
				if (ret[i].cost < ans.cost) {
					ans = ret[i];
				}
			}
		}
	}
	if (ans.cost != 0) {
		printf("%d", ans.cost);
	}
	else {
		printf("TLE");
	}

	return 0;
}