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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define len(v) ll(v.size())

template <class T>
void cout_vec(const vector<T> &vec){
  for(auto itr:vec) cout<<itr<<' ';
  cout<<'\n';
}

typedef pair<ll,ll> P;
const ll mod=1e9+7;

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    vector<P> a;
    rep(i,m){
      int x,y;
      cin>>x>>y;
      a.push_back(P(x,y));
      a.push_back(P(y,x));
    }
    rep(i,n){
      int ans=0;
      rep(j,2*m){
        if(a[j].first==i+1){
          ans++;
        }
      }
      cout<<ans<<endl;
    }
}