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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

struct segment_tree {
	struct modify {
		bool need;
		ll start;

		modify(bool need = false, ll start = 0) :
			need(need), start(start)
		{}
	};

	int n;
	ll add_all;
	vector<ll> tree;
	vector<modify> mod;

	segment_tree(int n = 0, ll val = 0) : n(n), add_all(0) {
		tree.resize(4 * n, val);
		mod.resize(4 * n);
	}

	void push(int v, int vl, int vr) {
		if (mod[v].need)
			tree[v] = mod[v].start;

		if (vr - vl > 1 && mod[v].need) {
			int vm = (vl + vr) / 2;
			mod[2 * v + 1].need = mod[2 * v + 2].need = true;
			mod[2 * v + 1].start = mod[v].start;
			mod[2 * v + 2].start = mod[v].start + vm - vl;
		}

		mod[v].need = false;
	}

	ll get(int L, int R) {
		return get(0, 0, n, L, R) + add_all;
	}

	ll get(int v, int vl, int vr, int L, int R) {
		push(v, vl, vr);
		if (R <= vl || vr <= L)
			return 1e18;
		if (L <= vl && vr <= R)
			return tree[v];

		int vm = (vl + vr) / 2;
		return min(get(2 * v + 1, vl, vm, L, R), get(2 * v + 2, vm, vr, L, R));
	}

	void update(int L, int R, ll start) {
		update(0, 0, n, L, R, start);
	}

	void update(int v, int vl, int vr, int L, int R, ll start) {
		push(v, vl, vr);
		if (R <= vl || vr <= L)
			return;
		if (L <= vl && vr <= R) {
			mod[v].need = true;
			mod[v].start = start + vl - L;
			return;
		}

		int vm = (vl + vr) / 2;
		update(2 * v + 1, vl, vm, L, R, start);
		update(2 * v + 2, vm, vr, L, R, start);

		push(2 * v + 1, vl, vm);
		push(2 * v + 2, vm, vr);
		tree[v] = min(tree[2 * v + 1], tree[2 * v + 2]);
	}
};

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	int n, m;
	cin >> n >> m;
	vector<int> a(n), b(n);
	for (int i = 0; i < n; i++)
		cin >> a[i] >> b[i], a[i]--;

	segment_tree dp(m, 0);
	for (int i = 0; i < n; i++) {
		dp.update(a[i], b[i], a[i] ? min(2ll * (n + m), dp.get(a[i] - 1, a[i])) + 1 : 2 * (n + m));
		dp.add_all++;

		int mn = dp.get(0, m);
		cout << (mn >= 2 * (n + m) ? -1 : mn) << '\n';
	}
}