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(){
  
  int N,T;
  scanf("%d%d",&N,&T);
  
  int i,c[N + 1],t[N + 1];
  for(i = 1; i <= N; i++){
    scanf("%d%d",&c[i],&t[i]);
  }
  
  int count = 0;
  for(i = 1; i <= N; i++){
    if(t[i] <= T)
      count++;
  }
  
  if(count == 0)
    printf("TLE");
  else{
    for(i = 1; i <= N; i++){
      if(t[i] > T)
        c[i] = 1001;
    }
    
    int ans = c[1];
    for(i = 2; i <= N; i++){
      if(ans > c[i])
        ans = c[i];
    }
    printf("%d",ans);
  }
  
  return 0;
  
}