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: Wrong Answer)

#include<stdio.h>
int main(void){
  int n,t,i,j,tmp;
  scanf("%d %d",&n,&t);
  int a[n][2];
  for(i=0;i<n;i++){
    scanf("%d %d",&a[i][0],&a[i][1]);
    if(a[i][1]>t){
      a[i][0]=1001;
    }
  }
  for(i=0;i<n;i++){
    for(j=i+1;j<n;j++){
      if(a[i][0]>a[j][0]){
        tmp=a[i][0];
        a[i][0]=a[j][0];
        a[j][0]=tmp;
      }
    }
  }
  if(a[0][1]<=t){
    printf("%d",a[0][0]);
  }else{
    printf("TLE");
  }
  return 0;
}