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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(){
        int N,M;
        scanf("%d %d", &N, &M);
        int a[100]={0},b[100]={0};
        int i,k;
        for(i=1;i<=M;i++){
                scanf("%d %d", &a[i], &b[i]);
        }
        int counta[100]={0},countb[100]={0};
        for(i=1;i<=M;i++){
                for(k=1;k<=N;k++){
                        if(a[i]==k){
                                counta[k]++;
                        }
                        if(b[i]==k){
                                countb[k]++;
                        }
                }
        }
        for(i=1;i<=N;i++){
                printf("%d\n",counta[i]+countb[i] );
        }
        return 0;
}