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,k=0,temp;
    scanf("%d %d",&n,&t);
    int ac[n],at[n],az[n],min=1001;
    for(int i=0; i<n; i++)
    {
        scanf("%d %d",&ac[i],&at[i]);
    }
    for(int j=0; j<n; j++)
    {
        if(at[j]> t)
        {
            ac[j]=1002;

        }
        if(ac[j]<min)
        {
            min=ac[j];
        }

    }

//    int min =1000;
//    for(int l=0; l<k; l++)
//    {
//        if(az[k]<min)
//        {
//            min=az[k];
//        }
//
//    }
    if(min!=1001)
    {
    printf("%d",min);
    }
    else
    {
        printf("TLE");
    }
    return 0;
}