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: Wrong Answer)

#include <stdio.h>

int min(int x, int y)
{
	if (x >= y) {
		return y;
	} else {
		return x;
	}
}

int main()
{
	int N, T, i,  min_c, flag, j;
	
	scanf("%d%d", &N, &T);
	
	int a[N][2];
	
	for (i = 0; i < N; i++) {
		a[i][1] = 0;
		a[i][0] = 0;
	}
	
	for (i = 0; i < N; i++) {
		scanf("%d%d", &a[i][0], &a[i][1]);
	}

	for (flag = 0, j = 0; flag == 0 && j < N; j++) {
		for (i = 0; i < N - 1; i++) {
			min_c = min(a[i][0], a[i + 1][0]);
		}
		for (i = 0; i < N; i++) {
			if (a[i][0] == min_c) {
				break;
			}
		}
		if (a[i][1] <= T) {
			printf("%d\n", a[i][0]);
			flag = 1;
		} else {
			a[i][0] = 1000;
			a[i][1] = 1000;
		}
	}
	
	if (j == N ) {
		printf("TLE\n");
	}
	
	return 0;
}