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, Ca=0, ir;
  scanf("%d", &N);
  scanf("%d", &T);
  int c[N];
  int t[N];
  for(i=0; i<N; i++){
    scanf("%d", &c[i]);
    scanf("%d", &t[i]);
    if (T < t[i]){
      t[i] = 1000;
      c[i] = 1000;
      Ca = Ca + 1;
    }
    else{
    }
  }
  for (i=0; i < N; i++){
    for (j=1; j<N; j++){
      if (c[i] > c[j]){
        ir = c[i];
        c[i] = c[j];
	c[j] = ir;
      }
      else{
      }
    }
  }
  if (Ca==N){
    printf("TLE");
  }
  else{
    printf("%d", c[0]);
  }
  return 0;
}