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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{	
	//input
	int N,T;
	scanf("%d %d",&N,&T);
	int c[N];
	int t[N];
	for (int i = 0; i < N; ++i)
	{
		scanf("%d %d",&c[i],&t[i]);
	}
	//main
	int p;
	int q;
	int M = N;
	for (int i = 0; i < N-1; ++i)
	{
		for (int j = 0; j < M-1; ++j)
		{
			if(c[j] > c[j+1]){
				p = c[j];
				q = c[j+1];
				c[j] = q;
				c[j+1] = p;
			}
		}
		M--;
	}

	for (int i = 0; i < N ; ++i)
	{
		if(i == N-1){
			printf("TLE");
			break;
		}
		if(t[i] <= T){
			printf("%d",c[i]);
			break;
		}
	}

	return 0;
}