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<bits/stdc++.h>
#define MAX(a,b) a>b?a:b
#define MIN(a,b) a<b?a:b
#define rep(i,n) for(int (i)=0;(i)<n;(i)++)
#define read(n) scanf("%d",&n);
#define readll(n) scanf("%lld",&n);
#define readdb(n) scanf("%lf",&n);
#define readch(n) scanf("%c",&n);
#define readstr(n) scanf("%s",n);
#define test printf("test\n");
#define init(n) memset(n,0,sizeof(n));

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;

int IT_MAX = 1 << 19;
int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const db PI = acos(-1);
const db ERR = 1e-10;
typedef struct p1{
	int first,second;	
}p; 
bool cmp (p a , p b){
	return a.first>b.first;
}
double qpow(double a, ll b) {
      double res = 1;
      while(b) {
          if (b & 1) res = res * a;
          a = a * a;
         b >>= 1;
     }
     return res;
}
ll C(ll n,ll x)
{
	ll ans = 1;
	for(ll i = n , j = 1 ; i >= n-x+1 ; i -- , j ++){
		ans *= i ;
		ans /= j ;
	}
	return ans;
}
int main(){
	int n , t;
	read(n);
	read(t);
	int mmin = (int)1e9;
	for(int i = 0 ; i < n ; i ++){
		int a , b;
		read(a);
		read(b);
		if(b <= t){
			mmin = min(mmin,a);
		}
	}
	if(mmin == 1e9) printf("TLE");
	else{
		printf("%d",mmin);
	}
	
	
	
	
	
	
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 return 0; 
}