#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define fix(x) fixed << setprecision(x)
#define eb emplace_back
constexpr char nl='\n';
using namespace std;
using ll = long long;
using ld = long double;
using vl = vector<long long>;
using vvl = vector<vector<long long>>;
using vs = vector<string>;
using pl = pair<long long, long long>;
template <typename T> inline bool chmin(T& a, const T& b) {bool compare = a > b; if (a > b) a = b; return compare;}
template <typename T> inline bool chmax(T& a, const T& b) {bool compare = a < b; if (a < b) a = b; return compare;}
template<class T>using rp_queue=priority_queue<T,vector<T>,greater<T>>;
void fast_io(){cin.tie(nullptr);ios_base::sync_with_stdio(false);}
template <typename T> T gcd(T a, T b) {if (b == 0)return a; else return gcd(b, a % b);}
template <typename T> inline T lcm(T a, T b) {return a /gcd(a, b)*b;}
const ld PI = acos(-1);
#include<atcoder/lazysegtree>
using S = long long;
using F = long long;

const S INF = 8e18;

S op(S a, S b){ return std::max(a, b); }
S e(){ return -INF; }
S mapping(F f, S x){ return f+x; }
F composition(F f, F g){ return f+g; }
F id(){ return 0; }
void solve(){
  int N,M;cin>>N>>M;
  std::vector<S> A(N+1);
  atcoder::lazy_segtree<S, op, e, F, mapping, composition, id> seg(A);
  //それぞれに何回マイナスその数がされたのかを求める配列
  //imosでやる
  vl minus(N+2),plus(N+2);
  //これを区間加算を使って高速化する
  while(M--){
    ll p,q;cin>>p>>q;
    for(int j=1;j<=N;j++){
      if(p-j>=0)seg.set(j,seg.get(j)+max(0LL,q-p+j));
      else seg.set(j,seg.get(j)+max(0LL,q+p-j));
    }
  }
  vl ans(N+1);
  for(int i=1;i<=N;i++){
    ans[i] = max(seg.get(i)+(plus[i]-minus[i])*i,0LL);
    cout<<ans[i]<<" ";
  }
  cout<<endl;
}
int main(){
  fast_io();
  int num_tc = 1;
  //cin >> num_tc;
  rep(tc,num_tc){
    //cout << "Case #" << tc+1 << ": " ;// << endl;
    solve();
  }
}