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 <bits/stdc++.h>

using namespace std;

#define int long long
#define pb push_back
#define FOR(i, a, b) for(int i=(a);i<(b);i++)
#define REP(i, n) FOR(i, 0, n)
#define RFOR(i, a, b) for(int i=(a);i>=(b);i--)
#define RREP(i, n) FOR(i, n, 0);

signed main () {
  cin.tie(0);
  cout << setprecision(10);

  int n, t; cin >> n >> t;
  vector<pair<int, int>> a(n);
  REP(i, n) {
    cin >> a[i].first >> a[i].second;
  }

  sort(a.begin(), a.end());

  REP(i, n) {
    if(a[i].second <= t) {
      cout << a[i].first << endl;
      return 0;
    }
  }

  cout << "TLE" << endl;
  
}