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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <list>
#include <algorithm>
#include <cmath>
#include <iomanip>
 
 
int main()
{
    int n, t;
    std::cin >> n >> t;
    std::vector<std::pair<int, int>> p;
    for(int i=0; i<n; i++)
    {
        int a, b;
        std::cin >> a >> b;
        p.push_back(std::make_pair(a, b));
    }
    std::sort(p.begin(), p.end());
    int ans = 10000;
    for(int i=0; i<n; i++)
    {
        if(p[i].second <= t)
        {
            ans = std::min(p[i].first, ans);
        }
    }
    if(ans==10000) std::cout << "TLE" << std::endl;
    else std::cout << ans << std::endl;
    return 0;
}