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 <iostream>
using namespace std;

long long h, w, st[200005 << 2], lazy[200005 << 2], inf = 0x3F3F3F3F3F;

void push(int x, int l, int mid){
    if (!lazy[x]) return;
    st[x<<1] = lazy[x];
    lazy[x<<1] = lazy[x];
    st[x<<1|1] = lazy[x]+(mid-l+1);
    lazy[x<<1|1] = lazy[x]+(mid-l+1);
    lazy[x] = 0;
}
void update(int x, int l, int r, int ul, int ur, long long u){
    if (ul <= l && r <= ur){
        st[x] = u;
        lazy[x] = u;
        return;
    }
    int mid = (l+r) >> 1;
    push(x, l, mid);
    if (ul <= mid) update(x<<1, l, mid, ul, ur, u);
    if (mid < ur) update(x<<1|1, mid+1, r, ul, ur, u+max(0, (mid-max(ul,l)+1)));
    st[x] = min(st[x<<1], st[x<<1|1]);
}
int query(int x, int l, int r, int q){
    if (q == 0) return inf;
    if (l == r) return st[x];
    int mid = (l+r) >> 1;
    push(x, l, mid);
    if (q <= mid) return query(x<<1, l, mid, q);
    else return query(x<<1|1, mid+1, r, q);
}

int main() {
    cin >> h >> w;
    for (int i = 1, a, b; i <= h; i++){
        cin >> a >> b;
        update(1, 1, w, a, b, query(1, 1, w, a-1)+1);
        if (st[1] > w) cout << -1 << "\n";
        else cout << st[1]+i << "\n";
    }
}