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>

int main(void)
{
  int n, t, i, j, k, h, tmp, tnp;

  scanf("%d %d", &n, &t);

  int c[n], s[n];

  for(h=0;h<n;h++){
    scanf("%d %d", &c[h], &s[h]);
  }

  for(i=0;i<n-1;i++){
    for(j=i+1;j<n;j++){
      if(c[i]>c[j]){
        tmp = c[i];
        c[i] = c[j];
        c[j] = tmp;

        tnp = s[i];
        s[i] = s[j];
        s[j] = tnp;
      }
    }
  }
int p;
  for(k=0;k<n;k++){
    if(s[k]<=t){
      p = k;
      break;}
      p = 100;
  }

  if(p == 100)
    printf("TLE");
  else
    printf("%d", c[p]);

  return 0;


}