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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

var (
	sc = bufio.NewScanner(os.Stdin)
)

func main() {
	sc.Split(bufio.ScanWords)
	n := nextInt()
	ds := make([]int, n)

	for i := 0; i < n; i++ {
		ds[i] = nextInt()
	}
	sort.Sort(sort.Reverse(sort.IntSlice(ds)))

	c := 0
	last := ds[0] + 1
	for _, d := range ds {
		if d < last {
			last = d
			c++
		}
	}
	fmt.Println(c)
}

func nextInt() int {
	i, err := strconv.Atoi(nextWord())
	pnc(err)
	return i
}

func nextWord() string {
	sc.Scan()
	return sc.Text()
}

func pnc(err error) {
	if err != nil {
		panic(err)
	}
}