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 JavaScript (node.js v5.12) (judge verdict: Accepted)

function gets(str){
  this._s = str.split("\n");
  this._y = 0;
};
gets.prototype = {
  a: function(f){
    var s = this._s, y = this._y, r;
    if(typeof s[y] === "string")s[y] = s[y].split(" ").reverse();
    r = s[y].pop();
    if(!s[y].length)this._y++;
    return f?r:+r;
  },
  l: function(f){
    var s = this._s[this._y++].split(" ");
    return f?s:s.map(a=>+a);
  },
  m: function(n,f){
    var s = this._s, y = this._y, r = [], i = 0;
    if(f)for(;i < n; i++)r[i] = s[y+i].split(" ");
    else for(;i < n; i++)r[i] = s[y+i].split(" ").map(a=>+a);
    this._y += n;
    return r;
  },
};
var o=new gets(require("fs").readFileSync("/dev/stdin","utf8"));

console.log(main());
function main(){
  var n = o.a();
  var t = o.a();
  var c = o.m(n);
  var ans = 9e9;
  for(var i = 0; i < n; i++){
    if(c[i][1] <= t && c[i][0] < ans)ans = c[i][0];
  }
  if(ans === 9e9)return "TLE";
  return ans;
}