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>
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define reps(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; --i)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; --i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define repc2(i, s, n) for (int i = (s); i <= (int)(n); i++)
#define inf 2e9
#define linf 9000000000000000000ll
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

int h, w;
vector<pair<int, int>> vp;

int main() {
    cin >> h >> w;
    vector<int> ans(h + 1, inf);
    vp.resize(h + 1);
    ans[0] = 0;
    reps(i, h) {
        int a, b;
        scanf("%d%d", &a, &b);
        vp[i] = make_pair(a, b);
    }
    int cur_pos = 1;
    typedef pair<ll, int> P;
    priority_queue<P, vector<P>, greater<P>> pq1, pq2;
    priority_queue<P, vector<P>, greater<P>> &pqn = pq1, &pqo = pq2;
    pqo.push(make_pair(0, 1));
    rep(i, h) {
        ans[i] = (pqo.empty()) ? inf : pqo.top().first;
        while (!pqo.empty()) {
            auto p = pqo.top();
            int ccos = p.first;
            int cpos = p.second;
            pqo.pop();
            if (cpos < vp[i].first || vp[i].second < cpos) {
                pqn.push(make_pair(ccos + 1, cpos));
            }
            if (cpos < vp[i].first) pqn.push(make_pair(ccos + 1 + vp[i].first - 1 + cpos, ccos + vp[i].first - cpos - 1));
            if (cpos < vp[i].second && vp[i].second + 1 <= w)
                pqn.push(make_pair(ccos + 1 + vp[i].second + 1 - ccos, ccos - cpos+ vp[i].second + 1));
        }
        swap(pqn, pqo);
    }
    reps(i, h) {
        if (ans[i] < inf) {
            printf("%d\n", ans[i]);
        } else {
            printf("%d\n", -1);
        }
    }
    return 0;
}