Problem: <image>

William owns a flat in central London. He decided to rent his flat out for the next n days to earn some money.

Since his flat is in the center of the city, he instantly got m offers in the form (l_i, r_i), which means that someone wants to book the flat from day l_i until day r_i inclusive. To avoid spending a lot of time figuring out whether it's profitable for him to accept an offer, William decided to develop an algorithm. The algorithm processes all offers as they arrive and will only accept offer i if the following two conditions are satisfied:

  * r_i - l_i + 1 ≥ x. 
  * None of the days between l_i and r_i are occupied by a previously accepted offer 



William isn't sure what value x should have and he asks you for help. For all x from 1 to n he wants you to calculate the total number of days for which the flat would be occupied if the corresponding value will be assigned to x.

Input

The first line contains two integers n and m (1 ≤ n ≤ 5 ⋅ 10^4, 1 ≤ m ≤ 10^5), which are the number of days and the number of offers, respectively.

Each of the next m lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n), which describe the i-th renting offer. All offers are given in chronological order.

Output

Print n integers. The number in i-th line must be equal to the number of days the flat would be occupied if the algorithm will use the value of x equal to i.

Example

Input


6 5
2 3
3 5
1 1
1 5
1 6


Output


3
2
3
5
5
6

Note

The description of segments from the first sample test for each x: 

  * x = 1 — algorithm will approve offers: 1 (2..3), 3 (1..1). The total number of days for which William's flat will be rented out is 3 
  * x = 2 — algorithm will approve offers: 1 (2..3). The total number of days for which William's flat will be rented out is 2 
  * x = 3 — algorithm will approve offers: 2 (3..5). The total number of days for which William's flat will be rented out is 3 
  * x = 4 — algorithm will approve offers: 4 (1..5). The total number of days for which William's flat will be rented out is 5 
  * x = 5 — algorithm will approve offers: 4 (1..5). The total number of days for which William's flat will be rented out is 5 
  * x = 6 — algorithm will approve offers: 5 (1..6). The total number of days for which William's flat will be rented out is 6 

Solution (in CPP):

Solution:
// --------------------------------------------------<TEMPLATE>--------------------------------------------------
// --------------------<optimizations>--------------------
/*
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
*/ 
// -------------------</optimizations>--------------------
 
// ---------------<Headers and namespaces>----------------
#include <bits/stdc++.h>
using namespace std;
 
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
*/
 
// ---------------</Headers and namespaces>---------------
 
// -----------------<Defines and typedefs>----------------
// typedef tree<int,null_type,less<int>,rb_tree_tag, 
// tree_order_statistics_node_update> indexed_set; // use less_equal for multiset
// order_of_key (val): returns the no. of values less than val
// find_by_order (k): returns the iterator to kth largest element.(0-based)
 
typedef long double LD;
typedef long long ll;
// #define int ll
#define pb push_back
#define mp make_pair
#define REP(i,n) for (int i = 0; i < n; i++)
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define REPD(i,n) for (int i = n-1; i >= 0; i--)
#define FORD(i,a,b) for (int i = a; i >= b; i--)
#define remax(a,b) a = max(a,b)
#define remin(a,b) a = min(a,b)
#define all(v) v.begin(),v.end()
typedef map<int,int> mii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
#define F first
#define S second
#define PQ(type) priority_queue<type>
#define PQD(type) priority_queue<type,vector<type>,greater<type> >
#define WL(t) while(t --)
#define SZ(x) ((int)(x).size())
#define runtime() ((double)clock() / CLOCKS_PER_SEC)
#define inrange(i,a,b) ((i>=min(a,b)) && (i<=max(a,b)))
 
/*
// -----<SCANF>-----
#define sfi(x) scanf("%d",&x);
#define sfi2(x,y) scanf("%d%d",&x,&y);
#define sfi3(x,y,z) scanf("%d%d%d",&x,&y,&z);
 
#define sfl(x) scanf("%lld",&x);
#define sfl2(x,y) scanf("%lld%lld",&x,&y);
#define sfl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z);
#define sfl4(x,y,z,x1) scanf("%lld%lld%lld%lld",&x,&y,&z,&x1);
#define sfl5(x,y,z,x1,y1) scanf("%lld%lld%lld%lld%lld",&x,&y,&z,&x1,&y1);
#define sfl6(x,y,z,x1,y1,z1) scanf("%lld%lld%lld%lld%lld%lld",&x,&y,&z,&x1,&y1,&z1);
 
#define sfs(x) scanf("%s",x);
#define sfs2(x,y) scanf("%s%s",x,y);
#define sfs3(x,y,z) scanf("%s%s%s",x,y,z);
// ----</SCANF>-----
 
// ----<PRINTF>-----
#define pfi(x) printf("%d\n",x);
#define pfi2(x,y) printf("%d %d\n",x,y);
#define pfi3(x,y,z) printf("%d %d %d\n",x,y,z);
 
#define pfl(x) printf("%lld\n",x);
#define pfl2(x,y) printf("%lld %lld\n",x,y);
#define pfl3(x,y,z) printf("%lld %lld %lld\n",x,y,z);
 
#define pfs(x) printf("%s\n",x);
#define pfs2(x,y) printf("%s %s\n",x,y);
#define pfs3(x,y,z) printf("%s %s %s\n",x,y,z);
 
#define pwe(x) printf("%lld ",x); // print without end
// ----</PRINTF>----
*/
#define FLSH fflush(stdout)
#define fileIO(name) \
    freopen(name".in", "r", stdin); \
    freopen(name".out", "w", stdout);
#define PRECISION(x) cout << fixed << setprecision(x); 
#define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
 
// ----------------</Defines and typedefs>----------------
 
// -------------------<Debugging stuff>-------------------
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif
 
// ------------------</Debugging stuff>-------------------
 
// ------------------------<Consts>-----------------------
const int MX = 30000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI=acos(-1);
 
#ifdef int
const int INF=1e16;
#else
const int INF=1e9;
#endif
 
const int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
// -----------------------</Consts>-----------------------
/*
// -------------------------<RNG>-------------------------
mt19937 RNG(chrono::steady_clock::now().time_since_epoch().count()); 
#define SHUF(v) shuffle(all(v), RNG);
// Use mt19937_64 for 64 bit random numbers.
 
// ------------------------</RNG>-------------------------
 
// ----------------------<MATH>---------------------------
template<typename T> T gcd(T a, T b){return(b?__gcd(a,b):a);}
template<typename T> T lcm(T a, T b){return(a*(b/gcd(a,b)));}
int add(int a, int b, int c = MOD){int res=a+b;return(res>=c?res-c:res);}
int mod_neg(int a, int b, int c = MOD){int res;if(abs(a-b)<c)res=a-b;else res=(a-b)%c;return(res<0?res+c:res);}
int mul(int a, int b, int c = MOD){ll res=(ll)a*b;return(res>=c?res%c:res);}
int muln(int a, int b, int c = MOD){ll res=(ll)a*b;return ((res%c)+c)%c;}
ll mulmod(ll a,ll b, ll m = MOD){ll q = (ll)(((LD)a*(LD)b)/(LD)m);ll r=a*b-q*m;if(r>m)r%=m;if(r<0)r+=m;return r;}
template<typename T>T expo(T e, T n){T x=1,p=e;while(n){if(n&1)x=x*p;p=p*p;n>>=1;}return x;}
template<typename T>T power(T e, T n, T m = MOD){T x=1,p=e;while(n){if(n&1)x=mul(x,p,m);p=mul(p,p,m);n>>=1;}return x;}
template<typename T>T extended_euclid(T a, T b, T &x, T &y){T xx=0,yy=1;y=0;x=1;while(b){T q=a/b,t=b;b=a%b;a=t;\
t=xx;xx=x-q*xx;x=t;t=yy;yy=y-q*yy;y=t;}return a;}
template<typename T>T mod_inverse(T a, T n = MOD){T x,y,z=0;T d=extended_euclid(a,n,x,y);return(d>1?-1:mod_neg(x,z,n));}
 
const int FACSZ = 1; // Make sure to change this
 
int fact[FACSZ],ifact[FACSZ];
 
void precom(int c = MOD){
    fact[0] = 1;
    FOR(i,1,FACSZ) fact[i] = mul(fact[i-1],i,c);
    ifact[FACSZ-1] = mod_inverse(fact[FACSZ-1],c);
    REPD(i,FACSZ-1){
        ifact[i] = mul(i+1,ifact[i+1],c);
    }
}
 
int ncr(int n,int r,int c = MOD){
    return mul(mul(ifact[r],ifact[n-r],c),fact[n],c);
} 
*/
// ----------------------</MATH>--------------------------
// --------------------------------------------------</TEMPLATE>--------------------------------------------------
 
void solvethetestcase();
 
signed main(){
    // comment when using scanf,printf
    FAST_IO
    int t = 1;
    // (UNCOMMENT FOR MULTIPLE TEST CASES)
    // cin >> t;
    FOR(testcase,1,t+1){
        // (UNCOMMENT FOR CODEJAM)
        // printf("Case #%lld: ",testcase); 
        solvethetestcase();
    }
}   
 
int n;
pii a[100005];
vi gg[100005];
int ans[100005];
int cnt;

int seg[MX],lft[MX],rgt[MX],ot[MX];
int sz = 1;

void upd_y(int ind,int l,int r,int x,int y,int val){
    if(l == r){
        remax(seg[ind],val);
        return;
    }
    int m = (l+r)/2;
    if(y <= m){
        if(!lft[ind]) lft[ind] = ++sz;
        upd_y(lft[ind],l,m,x,y,val);
    }
    else{
        if(!rgt[ind]) rgt[ind] = ++sz;
        upd_y(rgt[ind],m+1,r,x,y,val);
    }
    seg[ind] = max(seg[lft[ind]],seg[rgt[ind]]);
}

void upd_x(int ind,int l,int r,int x,int y,int val){
    if(l < r){
        int m = (l+r)/2;
        if(x <= m){
            if(!lft[ind]) lft[ind] = ++sz;
            upd_x(lft[ind],l,m,x,y,val);
        }
        else{
            if(!rgt[ind]) rgt[ind] = ++sz;
            upd_x(rgt[ind],m+1,r,x,y,val);
        }
    }
    if(!ot[ind]) ot[ind] = ++sz;
    upd_y(ot[ind],1,n,x,y,val);
}

int query_y(int ind,int l,int r,int x1,int x2,int y1,int y2){
    if(r < y1 or l > y2) return 0;
    if(l >= y1 and r <= y2){
        return seg[ind];
    }
    int m = (l+r)/2;
    return max(query_y(lft[ind],l,m,x1,x2,y1,y2),query_y(rgt[ind],m+1,r,x1,x2,y1,y2));
}

int query_x(int ind,int l,int r,int x1,int x2,int y1,int y2){
    if(r < x1 or l > x2) return 0;
    if(l >= x1 and r <= x2){
        if(!ot[ind]) return 0;
        return query_y(ot[ind],1,n,x1,x2,y1,y2);
    }
    int m = (l+r)/2;
    return max(query_x(lft[ind],l,m,x1,x2,y1,y2),query_x(rgt[ind],m+1,r,x1,x2,y1,y2));
}

void solve(int l,int r){
    if(l > r) return;
    int ind;
    ind = query_x(1,1,n,l,r,l,r);
    if(!ind) return;
    ind = 100001-ind;
    cnt += a[ind].S-a[ind].F+1;
    solve(l,a[ind].F-1);
    solve(a[ind].S+1,r);
}
 
void solvethetestcase(){
    int m;
    cin >> n >> m;
    FOR(i,1,m+1){
        cin >> a[i].F >> a[i].S;
        gg[a[i].S-a[i].F+1].pb(i);
    }
    FORD(x,n,1){
        for(auto i:gg[x]){
            upd_x(1,1,n,a[i].F,a[i].S,100001-i);
        }
        cnt = 0;
        solve(1,n);
        ans[x] = cnt;
    }
    FOR(i,1,n+1) cout << ans[i] << "\n";
}