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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <sstream>
#include <string>

using namespace std;

void solve(int N, int T, vector<pair<int, int>> X) {
  int cc = 1000;
  bool TLE = true;
  for (auto x : X) {
    if (x.first <= cc && x.second <= T) {
      cc = x.first;
      TLE = false;
    }
  }
  if (TLE) {
    printf("TLE\n");
  } else {
    printf("%d\n", cc);
  }
}

int main() {
  int N, T;
  cin >> N >> T;
  vector<pair<int, int>> X(N);
  for (int i = 0; i < N; i++) {
    int c, t;
    cin >> c >> t;
    X[i] = make_pair(c, t);
  }
  solve(N, T, X);
  return 0;
}