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 <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <deque>
using namespace std;
#define rep(i,n) for (ll i=0;i<n;i++)
typedef long long ll;

int N;
int T;
int c[110];
int t[110];

int main() {
	cin >> N >> T;
	rep(i, N) cin >> c[i] >> t[i];
	bool j = false;
	int ans = 0;
	rep(i, N) {
		if (!j && t[i] <= T) {
			ans = c[i];
			j = true;
		}
		else if (t[i] <= T) {
			ans = min(ans, c[i]);
		}
	}
	if (j) cout << ans << endl;
	else cout << "TLE" << endl;
	return 0;
}