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 <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <iterator>  // std::back_inserter()
#include <set>
#include <algorithm>  // std::copy()
#include <functional>  // std::greater<T>()
#include <utility>  // std::swap()
#include <numeric>  // accumulate(ALL(vec), 0)  0 は初期値
#include <cmath>
#include <climits>  // INT_MIN
#include <cctype>  // std::isdigit()
using namespace std;

#define ALL(obj) (obj).begin(), (obj).end()
#define REP(i, n) for (int i=0; i < (int)(n); i++)  // 0 ~ n-1
#define REPN(i, n) for (int i=1; i <= (int)(n); i++)  // 1 ~ n
#define MIN(vec) min_element(ALL((vec)))  // イテレータのため、値を取り出すときは * を先頭につける
#define MAX(vec) max_element(ALL((vec)))
#define IDX(vec, element_iter) distance((vec).begin(), element_iter)
#define SUM(vec) accumulate(ALL((vec)), 0)  // 0 は初期値
#define COPY(vec1, vec2) copy(ALL(vec1), back_inserter(vec2))  // vec1をvec2にコピーする vec2は空にしておく必要あり
typedef long long ll;

const int MOD = 1000000007;  // 1 000 000 007


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    int N, M;
    cin >> N >> M;
    vector<int> vec(2*M), count(60, 0);
    REP(i, 2*M) cin >> vec[i];
    
    // calculation
    REP(i, 2*M) {
        count[vec[i]]++;
    }
    
    REPN(i, N) {
        cout << count[i] << "\n";
    }
    
    return 0;
}