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 <math.h>
int main(void) {
	int N, T;
	scanf("%d%d", &N, &T);
	int c[100];
	int t[100];
	int c1[100];
	int t1[100];
	int count = 0;
	int min = 1000;
	int a = 0;
	for (int i = 0; i < N; i++) {
		scanf("%d%d", &c[i], &t[i]);
		if (T >= t[i]) {
			t1[count] = t[i];
			c1[count] = c[i];
			count++;
		}
	}


	if (count == 0) {
		printf("TLE");
	}
	else {

	for (int i = 0; i < count; i++) {
		if (c1[i] < min) {
			min = c1[i];
			a = i;
		}
	}

	printf("%d", c1[a]);

}



	return 0;

}