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>
typedef long long ll;
#define _GLIBCXX_DEBUG
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define all(v) v.begin(), v.end()
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using pii = pair<int, int>;
const int INF = 1 << 30;
const ll INFll = 1ll << 60;
template <typename T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}

int main() {
  int N, T;
  cin >> N >> T;
  vector<pii> P(N);
  rep(i, N) {
    int c, t;
    cin >> c >> t;
    P[i] = make_pair(c, t);
  }
  sort(all(P));
  pii ans;
  for (auto p : P) {
    if (p.second <= T) {
      cout << p.first << endl;
      return 0;
    }
  }
  cout << "TLE" << endl;

  return 0;
}