Problem:
Score : 600 points

Problem Statement

There is a grid of squares with H+1 horizontal rows and W vertical columns.

You will start at one of the squares in the top row and repeat moving one square right or down. However, for each integer i from 1 through H, you cannot move down from the A_i-th, (A_i + 1)-th, \ldots, B_i-th squares from the left in the i-th row from the top.

For each integer k from 1 through H, find the minimum number of moves needed to reach one of the squares in the (k+1)-th row from the top. (The starting square can be chosen individually for each case.) If, starting from any square in the top row, none of the squares in the (k+1)-th row can be reached, print -1 instead.

Constraints

1 \leq H,W \leq 2\times 10^5

1 \leq A_i \leq B_i \leq W

Input

Input is given from Standard Input in the following format:

H W
A_1 B_1
A_2 B_2
:
A_H B_H

Output

Print H lines. The i-th line should contain the answer for the case k=i.

Sample Input 1

4 4
2 4
1 1
2 3
2 4

Sample Output 1

1
3
6
-1

Let (i,j) denote the square at the i-th row from the top and j-th column from the left.

For k=1, we need one move such as (1,1) → (2,1).

For k=2, we need three moves such as (1,1) → (2,1) → (2,2) → (3,2).

For k=3, we need six moves such as (1,1) → (2,1) → (2,2) → (3,2) → (3,3) → (3,4) → (4,4).

For k=4, it is impossible to reach any square in the fifth row from the top.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int INF = 1e8;

struct LazySegmentTree {
    int size;
    vector<int> node, lazy;

    LazySegmentTree(int n) {
        size = 1;
        while (size < n) size <<= 1;
        node.resize(2 * size, INF);
        lazy.resize(2 * size, -1);
        for (int k = 1; k <= n; k++) {
            node[k + size] = 0;
        }
        for (int k = size - 1; k > 0; k--) {
            node[k] = min(node[2 * k], node[2 * k + 1]);
        }
    }

    void eval(int k, int l, int r) {
        if (lazy[k] == -1) return;
        if (k < size) { // not a leaf
            lazy[2 * k] = lazy[k];
            lazy[2 * k + 1] = lazy[k] + (r - l) / 2;
        }
        node[k] = lazy[k];
        lazy[k] = -1;
    }

    void update(int a, int b, int x, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;

        eval(k, l, r);
        if (r <= a || b <= l) return;
        if (a <= l && r <= b) {
            lazy[k] = x + l - a;
            eval(k, l, r);
        } else {
            int m = (l + r) / 2;
            update(a, b, x, 2 * k, l, m);
            update(a, b, x, 2 * k + 1, m, r);
            node[k] = min(node[2 * k], node[2 * k + 1]);
        }
    }

    int query(int a, int b, int k = 1, int l = 0, int r = -1) {
        if (r == -1) r = size;

        eval(k, l, r);
        if (r <= a || b <= l) return INF;
        if (a <= l && r <= b) return node[k];
        int m = (l + r) / 2;
        int vl = query(a, b, 2 * k, l, m);
        int vr = query(a, b, 2 * k + 1, m, r);
        return min(vl, vr);
    }
};

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

    int H, W;
    cin >> H >> W;
    LazySegmentTree st(W + 1);
    for (int i = 0; i < H; i++) {
        int A, B;
        cin >> A >> B;
        int x = st.query(A - 1, A);
        st.update(A, B + 1, x + 1);
        int ans = i + 1 + st.query(0, W + 1);
        if (ans >= INF) cout << -1 << "\n";
        else cout << ans << "\n";
    }
}