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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int>   II;
typedef vector< II >      VII;
typedef vector<int>     VI;
typedef vector< VI > 	VVI;
typedef long long int 	LL;
typedef vector<LL> 		VL;
typedef unsigned long long int ULL;

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define ALL(a) a.begin(),a.end()
#define SET(a,b) memset(a,b,sizeof(a))
#define LET(x,a) __typeof(a) x(a)

#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
//Works for forward as well as backward iteration

#define gu getchar
#define pu putchar
#define si(n) cin>>n
#define dout(n) cout<<n<<"\n"
//# define M_PI           3.14159265358979323846264338327950288

#define DRT() int t; si(t); while(t--)

#define PlUSWRAP(index,n) index = (index+1)%n		//index++; if(index>=n) index=0
#define MINUSWRAP(index,n) index = (index + n -1)%n 	//index--; if(index<0) index=n-1
#define ROUNDOFFINT(d) d = (int)((double)d + 0.5)	//Round off d to nearest integer

#define FLUSHN while(gu()!='\n')
#define FLUSHS while(gu()!=' ')

#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif

#define off ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)

// set_name.find_by_order(k) It returns to an iterator to the kth element (counting from zero) in the set in O(logn) time
// set_name.order_of_key(k) It returns to the number of items that are strictly smaller than our item k in O(logn) time.

int main() {
	#ifndef ONLINE_JUDGE
    freopen("../input.txt", "r", stdin);
    freopen("../output.txt", "w", stdout);
	#endif
	srand(chrono::steady_clock::now().time_since_epoch().count());
	off;
	LL n,t;
	cin>>n>>t;
	VL tt(n);
	VL cost(n);
	LL m = INT_MAX;
	rep(i,0,n){
		cin>>cost[i];
		cin>>tt[i];
		if(tt[i] <= t){
			m = min(m, cost[i]);
		}
	}
	cout<<m;
  	return 0;
}