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 <tourist>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> p;
const int INF = (1 << 28);
const int MOD = 1000000007;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define REP(i, n) for (int i = 0; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl; 
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
//setprecision(15)有効数字15桁
//-std=c++14
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b){return a * b / gcd(a, b);}
int n, m;
vector<int> a;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
  	int t;
    cin >> n>>t;
  	int mina=INF;
    for (int i = 0; i < n; i++)
    {
        int temp, temp1;
        cin >> temp>>temp1;
        if(temp1<=t) chmin(mina,temp);
    }
    if(mina==INF) return cout<<"TLE"<<"\n",0;
    cout << mina << "\n";
}