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)

//abc112_b.cpp
//Fri Nov  2 22:25:45 2018

#include <iostream>
#include <string>
#include <queue>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include <math.h>
#define INTINF 2147483647
#define LLINF 9223372036854775807
using namespace std;
using ll=long long;
typedef pair<int,int> P;

int main(){
	int n,T;
	cin >> n >> T;

	int c[n],t[n];
	int count = 0;
	for (int i=0;i<n;i++){
		cin >> c[i] >> t[i];
		if (t[i] > T){
			c[i] = INTINF;
		}else{
			count++;
		}
	}

	if (count == 0){
		cout << "TLE" << endl;
	}else {
		sort(c,c+n);
		cout << c[0] << endl;
	}

}