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 a,t,i=0,j=0,s=0,min;
   int b[100];
   int c[100];
   int k[100];
   int l[100];
   
   scanf("%d%d",&a,&t);
   while(a>0){
      scanf("%d%d",&b[i],&c[j]);
      a--;i++;j++;
   }
   j--;i--;
    while(j>=0){
        if(c[j]<=t){
            k[s]=c[j];
            l[s]=b[j];
            s++;
        }
        j--;
    }
    s--;
    min=l[s];
    while(s>=0){
        if(min>l[s]){
            min=l[s];
        }
        s--;
    }
    printf("%d",min);
    return 0;
   }