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>
#include <stdlib.h>
#include <limits.h>

int main(void)
{
	int n, T, err, i, min;
	int *c, *t;
	err = scanf("%d%d", &n, &T);
	if (err!=2 || (n<1||n>100) || (T<1||T>1000)) exit(1);
	c = (int *)malloc(sizeof(int)*n);
	t = (int *)malloc(sizeof(int)*n);
	for (int i = 0; i < n; ++i)
	{
		err = scanf("%d%d", &c[i], &t[i]);
		if (err!=2 || (c[i]<1||c[i]>1000) || (t[i]<1||t[i]>1000)) exit(1);
	}

	for (int i = 0; i < n; ++i)
	{
		if (t[i] > T) c[i] = INT_MAX;
	}
	min = c[0];
	for (int i = 1; i <= n-1; ++i)
	{
		if (c[i] < min) min = c[i];
	}
	if (min < INT_MAX)
	{
		printf("%d\n", min);
	}else
	{
		printf("TLE\n");
	}
	return 0;
}