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>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INT_MAX 2147483647

int main(void){
    int N, T;
    scanf("%d%d", &N, &T);

    int *c, *t;
    c = (int*)malloc(N * sizeof(int));
    t = (int*)malloc(N * sizeof(int));

    int i;
    int min = INT_MAX;
    for(i=0; i<N; i++) {
        scanf("%d%d", &c[i], &t[i]);
        if(t[i] <= T) {
            if(c[i] < min) {
                min = c[i];
            }
        }
    }
    if(min != INT_MAX) {
        printf("%d\n", min);
    }
    else{
        printf("TLE\n");
    }

    free(c);
    free(t);

    return 0;
}