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<iostream>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include <algorithm>
#include <set>
#include<math.h>
#include<cmath>
#include<vector>
#include<map>
#include<random>
#include<sstream>
using namespace std;

int main(){
    int N,T;cin >> N >> T;
    int c[1000],t[1000] = {};
    int f = 0;
    int m = 1000;
    int mId = 0;
    for(int i = 0;i < N;i++){
        cin >> c[i] >> t[i];
        if(T < t[i])f = 1;
        if(m > t[i] && t[i] >= T)m =t[i];mId = c[i];
    }
    if(f == 0)cout << "TLE" << endl;
    else cout << mId << endl;

    

}