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 <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

int n, t;
int ans = 199999999;

struct node{
	int ti;
	int ci;
	bool operator < (const node & a) const
	{
		if (a.ti == ti) return ci < a.ci;
		return  ti < a.ti;
	}
}a[3333];

int main()
{
	cin >> n >> t;
	for (int i = 1; i <= n; ++i) cin >> a[i].ci >> a[i].ti;
	sort(a + 1, a + n + 1);
	for (int i = 1; i <= n; ++i)
	{
		if (a[i].ti > t)  break;
		ans = min(a[i].ci, ans);
	}
	if (ans == 199999999) cout << "TLE";
	else cout << ans;
}