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<algorithm>
#include<set>
using namespace std;
#define MOD 200001
set<int> distFirst;
set<int> locFirst;
int main()
{
    int H,W;
    scanf("%d %d", &H, &W);
    for(int i = 0; i < W; i ++)
    {
        distFirst.insert(i);
        locFirst.insert(i * MOD + i);
    }
    for(int i = 0; i < H; i ++)
    {
        int l,r;
        scanf("%d %d", &l, &r);
        l -= 1;
        r -= 1;
        set<int>::iterator firstOne = locFirst.lower_bound(l * MOD);
        set<int>::iterator last = locFirst.end();
        while(true)
        {
            if(next(firstOne) != last && *next(firstOne) < ((r + 1) * MOD))
            {
                int deletableLoc = *firstOne / MOD;
                int deletablePrevLoc = *firstOne % MOD;
                int deletableDist = deletableLoc - deletablePrevLoc;
                ++firstOne;
                locFirst.erase(deletableLoc * MOD + deletablePrevLoc);
                distFirst.erase(deletableDist * MOD + deletableLoc);
            }
            else
            {
                int deletableLoc = *firstOne / MOD;
                int deletablePrevLoc = *firstOne % MOD;
                int deletableDist = *firstOne % MOD;
                int newLoc = r + 1;
                locFirst.erase(deletableLoc * MOD + newLoc);
                distFirst.erase((newLoc - deletablePrevLoc) * MOD + deletableLoc);
                if(r != W - 1)
                {
                    locFirst.insert((r + 1) * MOD + deletablePrevLoc);
                    distFirst.insert((newLoc - deletablePrevLoc) * MOD + r + 1);
                }
                break;
            }
        }
        if(distFirst.empty())
        {
            printf("-1\n");
        }
        else
        {
        	int ans = (*distFirst.begin()) / MOD;
            printf("%d\n",ans + i + 1);
        }
    }
}