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>
 
int main()
{
  int a, b, c, d, X, Y;
  int D, G;
  //char s[101][101]={};
  char sv[101]={};
  char ss[101]={};
//  char s_temp[101]={};
  char t[101]={};
//  char s[10000000000000000001]={};
  char s[2][100001]={};
  int len=0;
  int N, M;
//  int A[1000000001]={};
  int A[1001]={};
//  long int A[1001]={};
  int B[1001]={};
  int C[20000]={};
  int i,j,k;
  int flag=0;
  int tmp=0;
  int min=0;
  int cnt=0;
  int ans=0;
  int max=0;
  int max_o=0;
  int max2_o=0;
  int max_o_num=0;
  int max_e=0;
  int max2_e=0;
  int max_e_num=0;
//  int ans=0;
  int str_len=0;
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
  int ans_o[100001]={};
  int ans_e[100001]={};
  min=1001;
  scanf("%d %d", &a, &b);
  for(i=0;i<a;i++){
    scanf("%d %d", &A[i], &B[i]);
    if(min>=A[i]){
      if(b>=B[i]){
        min=A[i];
        flag=1;
      }
    }
  }
  if(flag==0)   printf("TLE\n");
  else          printf("%d\n",min);
  
return 0;
}