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: Wrong Answer)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<map>
#include<cstdio>
#include<stack>
#include<queue>
#include<fstream>
#include<cstdio>
#include<deque>
#include<iomanip>
#include <numeric>
//#include<bits/stdc++.h>

using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=(s);i<(n);i++)
#define repr(i,n) for(int i=n-1;i>=0;i--)
#define REPR(i,s,n) for(int i=(s);i>=(n);(i)--)
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
const double pi = acos(-1.0);

typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
#define mod 1e9 + 7;
ll gcd(ll x, ll y)
{
	ll r;
	while ((r = x % y) != 0)
	{
		x = y;
		y = r;
	}
	return y;
}
ll lcm(ll x, ll y)
{
	x /= gcd(x, y);
	y /= gcd(x, y);
	return (x * y);
}


int main() {
	ll n, t,ret=1001;
	cin >> n >> t;
	rep(i, n) {
		ll c1, t1;
		cin >> c1 >> t1;
		if (t1 <= t && ret > c1) {
			ret = c1;
		}
	}
	if (ret == 1001) cout << "TLE" << endl;
	cout << ret << endl;
}