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>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define sz(x) (int)(x).size()
#define uni(x) x.erase(unique(rng(x)),x.end())
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
#define EPS (1e-10)
#define equals(a, b) (fabs((a)-(b)) < EPS)
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef set<int> S;
typedef queue<int> Q;
typedef queue<P> QP;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
const int MOD = 1000000007;


  
int main() {
  int N,M;
  cin >> N >> M;
  vi cnt(N,0);
  rep(i,M) {
    int a,b;
    cin >> a >> b;
    a--;b--;
    cnt[a]++;
    cnt[b]++;
  }
  rep(i,N) {
    cout << cnt[i] << endl;
  }
  return 0;
}