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 int LL;
using namespace std;
#define INF (1e9)
#define LINF (1LL << 60)
#define ALL(x) (x).begin(),(x).end()
#define FOR(i,k,n) for(int i=(k);(i)<(n);(i)++)
#define REP(i,n) for(int i = 0;(i)<(n);(i)++)
#define REPR(i,n) for(int i = n; i >= 0; i--)
#define F first
#define S second
#define MOD 1000000007

int main() {
	int n, T;
	cin >> n >> T;
	vector<int> c(n), t(n);
	REP(i, n) cin >> c[i] >> t[i];
	int m = INF;
	REP(i, n) {
		if (t[i] > T) continue;
		m = min(m, c[i]);
	}
	if (m == INF) {
		cout << "TLE" << endl;
	}
	else {
		cout << m << endl;
	}
}