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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include <string.h>
#include <algorithm>
using namespace std;

int main()
{
    int n,t;
    int f = 1000;
    cin >> n >> t;
    int c[100],s[100];
    memset(c,0,sizeof(c));
    memset(s,0,sizeof(s));
    
    for(int i = 0;i < n;i++)
    {
        int j,k;
        cin >> j >> k;
        c[i] = j;
        s[i] = k;
     }
     
     for(int j= 0;j < n;j++)
     {
        if(s[j] <= t)
        {
            if(c[j] <= f && c[j] > 0)
            {
                f = c[j];
            }
        }
     }
    if(f == 1000)
    {
        cout << "TLE";
    }else
    {
        cout << f;
    }
    return 0;
}