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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#include <malloc.h>

#define MAX 9999

int Pow(int n, int m);
int Max(int N[], int n);
int Min(int N[], int n);
int Sum(int N[], int n);

void Move(int* x, int* y, int d, int n);
void Upper(int* x, int* y, int d);
void Lower(int* x, int* y, int d);
void Right(int* x, int* y, int d);
void Left(int* x, int* y, int d);




int main()
{
	
	int N = 0, T = 0;
	int c[100] = { MAX };
	int t[100] = { MAX };
	
	scanf("%d%d", &N, &T);
	
	for (int i = 0; i < N; i++) {
		scanf("%d%d", &c[i], &t[i]);
	}

	for (int i = 0; i < N; i++) {
		if (t[i] > T) {
			c[i] = MAX;
		}
	}

	int ans = Min(c, N);

	if (ans == MAX) {
		printf("TLE");
		return 0;
	}

	printf("%d", ans);

	



	return 0;
}

void Move(int* x, int* y, int d, int n) {

}

void Upper(int* x, int* y, int d) {
	*y += d;
}
void Lower(int* x, int* y, int d) {
	*y -= d;
}
void Right(int* x, int* y, int d) {
	*x += d;
}
void Left(int* x, int* y, int d) {
	*x -= d;
}

int Pow(int n, int m) {
	int ret = 1;

	for (int i = 0; i < m; i++) {
		ret *= n;
	}

	return ret;
}

int Max(int N[], int n) {
	int max = N[0];

	for (int i = 0; i < n; i++) {
		max = max < N[i] ? N[i]: max;
	}

	return max;
}

int Min(int N[], int n) {
	int min = N[0];

	for (int i = 0; i < n; i++) {
		min = min > N[i] ? N[i] : min;
	}

	return min;
}

int Sum(int N[], int n) {
	int sum = 0;

	for (int i = 0; i < n; i++) {
		sum += N[i];
	}

	return sum;
}