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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define SPEED ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define int long long
#define ld long double
#define fi first
#define se second
#define all(uiet) uiet.begin(),uiet.end()
#define read(UIET) for(int i = 0; i < n; ++i) cin >> UIET[i]
#define out(UIET) for(int i = 0; i < n; ++i) cout << UIET[i]
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define vpp vector<pair< int, int > >
#define pll pair<int , int >
#define ppll pair < pll , pll >
#define debug(n1) cout << n1 << endl
#define len(a) ((int) (a).size())
#define endl "\n"
#define mod 1000000007
const int INF=(1LL<<60)-1;
const int MAXN = 1000001;
vector<int> adj[MAXN];
int power(int x, int y, int p = mod){ 
    int res = 1; 
    x = x % p; 
    if (x == 0) return 0; 
    while (y > 0) {  
        if (y & 1) res = (res*x) % p; 
        y = y >> 1;
        x = (x * x) % p;   
    } 
    return res; 
} 
int modFact(int n, int p = mod) 
{ 
    if (n >= p) 
        return 0; 
  
    int result = 1; 
    for (int i = 1; i <= n; i++) 
        result = (result * i) % p; 
  
    return result; 
} 
int32_t main(){
    SPEED;
    int n, m;
    cin >> n >> m;
    for(int i = 0; i < m; ++i){
        int u, v;
        cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    for(int i = 1; i <= n; ++i){
        cout << adj[i].size() << endl;
    }
}
// check indices of loop
// check for overflow