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,i,j=0,min=99999;
    scanf("%d %d",&n,&t);

    int cost[n],time[n],cost2[n],time2[n];

    for(i=0;i<n;i++)
    {
        scanf("%d %d",&cost[i],&time[i]);
    }

      for(i=0;i<n;i++)
    {
        cost2[i]=0;
    }

    for(i=0;i<n;i++)
    {
        if(time[i]<=t)
        {
            cost2[j]=cost[i];
            time2[j]=time[i];
            j++;
        }
    }

    for(i=0;i<j;i++)
    {
        if(cost2[i]<=min)min=cost2[i];
    }

    if(min==99999)printf("TLE");
    else
    printf("%d",min);




}