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;
struct play{
	int c,t;
};play h[1010];
bool cmp1(play a,play b)
{
	return a.t<b.t;
}
bool cmp2(play a,play b)
{
	return a.c<b.c;
}
int main(){
	int n,m,cnt=0,top=1,cost=0;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&h[i].c,&h[i].t);
	}
	sort(h+1,h+1+n,cmp1);
	sort(h+1,h+1+n,cmp2);
	while(7)
	{
		if(h[top].t>m)top++;
		else
		{
			printf("%d\n",h[top].c);
			return 0;
		}
	}
	return 0;
}