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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;

static const double EPS = 1e-8;
static const double PI = 4.0 * atan(1.0);
static const ll INF = 1023456789;

#define REP(i,n)	for(int i=0;i<n;++i)
#define REPR(i,n)	for(int i=n-1;i>=0;--i)
#define FOR(i,s,n)	for(int i=s;i<n;++i)
#define FORR(i,s,n)	for(int i=n-1;i>=s;--i)
#define ALL(c)		(c).begin(),(c).end()
#define CLEAR(v)	memset(v,0,sizeof(v))
#define MP(a,b)		make_pair((a),(b))
#define ABS(a)		((a)>0?(a):-(a))
#define F			first
#define S			second

int main(int argc, char **argv) {
	ll n, lt, c, t, res = INF;
	cin >> n >> lt;
	REP(i, n) {
		cin >> c >> t;
		if (t <= lt) res = min(res, c);
	}
	if (res >= INF) cout << "TLE" << endl;
	else cout << res << endl;
	return 0;
}