Problem:
Score : 200 points

Problem Statement

An X-layered kagami mochi (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi.

Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?

Constraints

1 ≤ N ≤ 100

1 ≤ d_i ≤ 100

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
d_1
:
d_N

Output

Print the maximum number of layers in a kagami mochi that can be made.

Sample Input 1

4
10
8
8
6

Sample Output 1

3

If we stack the mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, we have a 3-layered kagami mochi, which is the maximum number of layers.

Sample Input 2

3
15
15
15

Sample Output 2

1

When all the mochi have the same diameter, we can only have a 1-layered kagami mochi.

Sample Input 3

7
50
30
50
100
50
80
30

Sample Output 3

4

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

// inputに入力データ全体が入る
function Main(input) {
	var input = input.split('\n')
	var N = parseInt(input[0])
	var d = input.slice(1)
	d = d.map(function (x) {
	    return parseInt(x, 10); 
	});

	//10, 9, 8...
	d.sort(function(a,b){
        if( a < b ) return 1;
        if( a > b ) return -1;
        return 0;
	});

	d = d.filter(function(item, pos) {
    	return d.indexOf(item) == pos;
	})

	var lastMochi = d[0],
		c = 1	
	for(var i = 1; i < d.length; i++) {
		var mochi = d[i]
		if(mochi < lastMochi) c++
		else break
	}

	console.log("" + c)
}

//*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));