#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define int long long int

template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}

typedef pair<int, int> pii;
typedef long long ll;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
constexpr ll INF = 1001001001001001LL;
constexpr ll MOD = 1000000007LL;

int N, M;
int rb[100010], lb[200010], ub[200010];

signed main() {
    cin >> N >> M;
    memset(rb, -1, sizeof(rb));
    iota(lb, lb+N, 0);
    iota(ub, ub+N, 1);
    rep(i,0,M) {
        int x, y; cin >> x >> y;
        rb[x] = y-1;
    }

    rep(i,0,100000) {
        if(rb[i] == -1) continue;
        int v = rb[i];
        int nl = min(lb[v], lb[v+1]);
        int nr = max(ub[v], ub[v+1]);

        lb[v] = lb[v+1] = nl;
        ub[v] = ub[v+1] = nr;
    }

    rep(i,0,N) {
        if(i == 0) printf("%lld", ub[i] - lb[i]);
        else printf(" %lld", ub[i] - lb[i]);
    }
    cout << endl;
    return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

Example

Input

4 3
1000 1
2000 2
3000 3


Output

2 3 4 4

### Response

```cpp
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <functional>
using namespace std;

#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define repq(i,a,n) for(int (i)=(a); (i)<=(n); (i)++)
#define repr(i,a,n) for(int (i)=(a); (i)>=(n); (i)--)
#define int long long int

template<typename T> void chmax(T &a, T b) {a = max(a, b);}
template<typename T> void chmin(T &a, T b) {a = min(a, b);}
template<typename T> void chadd(T &a, T b) {a = a + b;}

typedef pair<int, int> pii;
typedef long long ll;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
constexpr ll INF = 1001001001001001LL;
constexpr ll MOD = 1000000007LL;

int N, M;
int rb[100010], lb[200010], ub[200010];

signed main() {
    cin >> N >> M;
    memset(rb, -1, sizeof(rb));
    iota(lb, lb+N, 0);
    iota(ub, ub+N, 1);
    rep(i,0,M) {
        int x, y; cin >> x >> y;
        rb[x] = y-1;
    }

    rep(i,0,100000) {
        if(rb[i] == -1) continue;
        int v = rb[i];
        int nl = min(lb[v], lb[v+1]);
        int nr = max(ub[v], ub[v+1]);

        lb[v] = lb[v+1] = nl;
        ub[v] = ub[v+1] = nr;
    }

    rep(i,0,N) {
        if(i == 0) printf("%lld", ub[i] - lb[i]);
        else printf(" %lld", ub[i] - lb[i]);
    }
    cout << endl;
    return 0;
}
```