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>
int main()
{
	int i,N,T;
	int min, sum = 0;
	// コスト
	int* c;
	//　時間
	int* t;
	int* count;
	scanf("%d", &N);	scanf("%d", &T);

	// 配列の動的確保
	c = (int*)malloc(N*sizeof(int));
	t = (int*)malloc(N*sizeof(int));
	count = (int*)malloc(N*sizeof(int));

	for(i=0; i<N; i++){
		if(t[i]>T) count[i] = 0;
	}

	for(i=0; i<N; i++){
		scanf("%d", &c[i]); 
		scanf("%d", &t[i]);
		if(t[i]>T) count[i] = 1;
	}

	for(i=0; i<N; i++){
		sum += count[i];
	}

	min = c[0];

	for(i=1; i<N; i++){
		if(count[i] != 1){
			if(min > c[i]) min = c[i];
		}
	}

	if(sum==N) printf("TLE");
	else printf("%d", min);

	free(c);
	free(t);
	free(count);

	return 0;
}