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: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;

namespace ABC061
{
    class B
    {
        static void Main(string[] args)
        {

            int[] input = Console.ReadLine().Split().Select(int.Parse).ToArray();

            int N = input[0];
            int M = input[1];

            Dictionary<int, int> dict = new Dictionary<int, int>();

            int a = 0;
            int b = 0;

            for (int i = 0; i < M; i++)
            {
                input = Console.ReadLine().Split().Select(int.Parse).ToArray();

                a = input[0];
                b = input[1];

                if (!dict.ContainsKey(a)) { dict.Add(a, 0); }
                if (!dict.ContainsKey(b)) { dict.Add(b, 0); }

                if (dict.ContainsKey(a)) { dict[a]++; }
                if (dict.ContainsKey(b)) { dict[b]++; }

            }

            var res = dict.OrderBy(x => x.Key);

            foreach (var item in res)
            {
                Console.WriteLine(item.Value);
            }
        }
    }
}