Problem:
Score : 200 points

Problem Statement

There are N cities and M roads.
The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally.
There may be more than one road that connects the same pair of two cities.
For each city, how many roads are connected to the city?

Constraints

2≤N,M≤50

1≤a_i,b_i≤N

a_i ≠ b_i

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_M b_M

Output

Print the answer in N lines.
In the i-th line (1≤i≤N), print the number of roads connected to city i.

Sample Input 1

4 3
1 2
2 3
1 4

Sample Output 1

2
2
1
1

City 1 is connected to the 1-st and 3-rd roads.

City 2 is connected to the 1-st and 2-nd roads.

City 3 is connected to the 2-nd road.

City 4 is connected to the 3-rd road.

Sample Input 2

2 5
1 2
2 1
1 2
2 1
1 2

Sample Output 2

5
5

Sample Input 3

8 8
1 2
3 4
1 5
2 8
3 7
5 2
4 1
6 8

Sample Output 3

3
3
2
2
2
1
1
2

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Globalization;

class Myon
{
    static Scanner cin;
    public Myon() { }
    public static int Main()
    {
        //Console.SetOut(new Printer(Console.OpenStandardOutput()));
        cin = new Scanner();
        new Myon().calc();
        return 0;
    }


    public void calc()
    {
        int n = cin.nextInt(), m = cin.nextInt();
        int[] a = new int[m], b = new int[m];
        for(int i = 0;i<m;i++) { a[i] = cin.nextInt();b[i] = cin.nextInt(); }

        for(int i= 0; i < n; i++)
        {
            int count = 0;
            for(int j = 0; j < m; j++)
            {
                if (a[j] == i + 1 || b[j] == i + 1) count++;
            }
            ShowL($"{count}");
        }

    }

    void ShowL(string s)
    {
        Console.WriteLine(s);
    }
    
}

class Printer : StreamWriter
{
    public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
    public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    public Printer(Stream stream, Encoding encoding) : base(stream, encoding) { }
}

class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        while (i >= s.Length)
        {
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
        }
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }

}