Problem:
Score : 300 points

Problem Statement

There are N cities on a 2D plane. The coordinate of the i-th city is (x_i, y_i). Here (x_1, x_2, \dots, x_N) and (y_1, y_2, \dots, y_N) are both permuations of (1, 2, \dots, N).

For each k = 1,2,\dots,N, find the answer to the following question:

Rng is in City k.
Rng can perform the following move arbitrarily many times:

move to another city that has a smaller x-coordinate and a smaller y-coordinate, or a larger x-coordinate and a larger y-coordinate, than the city he is currently in.

How many cities (including City k) are reachable from City k?

Constraints

1 \leq N \leq 200,000

(x_1, x_2, \dots, x_N) is a permutation of (1, 2, \dots, N).

(y_1, y_2, \dots, y_N) is a permutation of (1, 2, \dots, N).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print N lines. In i-th line print the answer to the question when k = i.

Sample Input 1

4
1 4
2 3
3 1
4 2

Sample Output 1

1
1
2
2

Rng can reach City 4 from City 3, or conversely City 3 from City 4.

Sample Input 2

7
6 4
4 3
3 5
7 1
2 7
5 2
1 6

Sample Output 2

3
3
1
1
2
3
2

Solution in C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
#define show(x) {for(auto i: x){cout << i << " ";} cout << endl;}
using namespace std;
using ll = long long;
using P = pair<int, int>;
#include <atcoder/all>
using namespace atcoder;

template<typename T>
struct BIT {
    int n;
    vector<T> d;
    BIT(int n=0):n(n), d(n+1) {}
    void add(int i, T x=1) {
        for (i++; i <= n; i += i&-i) {
            //i&-iで最下位bitを足している
            d[i] += x;
        }
    }
    T sum(int i) {
        T x = 0;
        for (i++; i; i -= i&-i) {
            x += d[i];
        }
        return x;
    }
    T semiopenSum(int l, int r) {
        // [l,r)の半開区間
        return sum(r-1) - sum(l-1);
    }
};  // BIT<int> bit(N);

int op(int a, int b) {
    return max(a, b);
}

int e() {
    return (int)(0);
}

int main() {
    // BIT
    int N;
    cin >> N;
    vector<P> X(N);
    vector<int> Y(N);
    rep(i, N) {
        cin >> X[i].first >> Y[i];
        X[i].first--;
        Y[i]--;
        X[i].second = i;
    }
    sort(X.begin(), X.end());
    vector<int> cnt(N, 1);
    BIT<int> bit(N);
    for (int i = 0; i < N; i++) {
        int id = X[i].second;
        int y = Y[id];
        int down = bit.sum(y);
        cnt[id] += down;
        bit.add(y);
    }
    bit = BIT<int> (N);
    for (int i = N-1; i >= 0; i--) {
        int id = X[i].second;
        int y = Y[id];
        int down = bit.sum(y);
        cnt[id] += (N-1 - i) - down;
        bit.add(y);
    }
    rep(i, 15) {
        segtree<int, op, e> seg(N);
        for (int i = 0; i < N; i++) {
            int id = X[i].second;
            int y = Y[id];
            int c = cnt[id];
            int x = seg.prod(0, y);
            cnt[id] = max(cnt[id], x);
            seg.set(y, c);
        }
        segtree<int, op, e> seg2(N);
        for (int i = N-1; i >= 0; i--) {
            int id = X[i].second;
            int y = Y[id];
            int c = cnt[id];
            int x = seg2.prod(y, N+1);
            cnt[id] = max(cnt[id], x);
            seg2.set(y, c);
        }
    }

    rep(i, N) {
        cout << cnt[i] << '\n';
    }


    return 0;
    // ll, 0, -, 1i, for s&g, debug
}