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<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define YES cout<<"YES"<<endl
#define NO cout<<"NO"<<endl
#define Yes cout<<"Yes"<<endl
#define No cout<<"No"<<endl
#define ANS cout << ans << endl
#define all(a) a.begin(),a.end()
#define SORT(a) sort(all(a))
#define REVERSE(a) reverse(all(a))
#define int long long
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define VIin(a) for(int i=0; i<a.size(); i++) cin >> a[i]
#define OUT(a) cout << a << endl
typedef vector<int> VI;
typedef vector<VI> VVI;
const int MOD=1e9+7;
const long long INF = 1LL << 60;
/*最大公約数*/int gcd(int x, int y) { return y ? gcd(y,x%y) : x;}
/*最小公倍数*/int lcm(int a, int b) { int g = gcd(a, b); return a / g * b; }
/*素数判定*/int isPrime(int n){if(n<2){return 0;}for(int i=2; i*i<=n; i++){if(n%i==0){return 0;}}return 1;}
/*約数列挙*/vector<int> getDivisor(int n){vector<int> v; for(int i=1; i*i<=n; i++){if(n%i==0){v.push_back(i);if(i!=n/i){v.push_back(n/i);}}}sort(v.begin(),v.end()); return v;}
//----------------------------------------------
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n,t,flag = INF,ans; cin >> n >> t;
    rep(i,n){
        int c,time; cin >> c >> time;
        if(t > time) if(flag > time) ans = c , flag = time;
    }
    if(flag == INF){
        OUT("TLE");
        return 0;
    }
    ANS;

    return 0;
}