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 <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define rep(i, n) for (int i = 0; i < (n); i++)
#define reps(i, n, s) for (int i = (s); i < (n); i++)
#define MOD 1000000007
typedef long long ll;
using namespace std;
using Graph = vector<vector<int>>;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N, T;
    cin >> N >> T;
    int c, t;
    int ans=-1;
    int min_time = 1e9;
    rep(i, N)
    {
        cin >> c >> t;
        if (t<=T)
        {
            if(ans>c){
                ans = c;
            }
        }
    }
    if (ans <0)
    {
        cout << "TLE"<< endl;
    }
    else
    {
        cout << ans  << endl;
    }

    //system("pause");
    return 0;
}