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 <vector>
#include <string>
#include <algorithm>
#define PB push_back
#define SZ size()
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define GET(a) cin >> (a)
#define GET2(a, b) cin >> a >> b
#define GET3(a, b, c) cin >> a >> b >> c
#define SHOW(v) cout << v << endl
#define MOD 1000000007
typedef long long ll;

using namespace std;

struct xyz {
	int a;
	int b;
};

int main()
{
	int n, t;
	GET2(n, t);
	
	vector<xyz> v(n);
	REP(i, n) {
		GET2(v[i].a, v[i].b);
	}
	
	int ans = 100000;
	REP(i, n) {
		if(v[i].b <= t && ans > v[i].a) {
			ans = v[i].a;
		}
	}
	
	if(ans != 100000) {
		SHOW(ans);
	}
	else {
		SHOW("TLE");
	}
    return 0;
}