Problem: 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.

Example

Input

4 4
2 4
1 1
2 3
2 4


Output

1
3
6
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main(void){
    int h, w;
    cin >> h >> w;
    map<int, int> cur2start;
    multiset<int> values;

    for (int i = 0; i < w; i++) {
        cur2start[i] = i;
        values.insert(0);
    }

    for (int i = 0; i < h; i++) {
        int left, right;
        cin >> left >> right;
        auto it = cur2start.lower_bound(left - 1);
        int right_bound_of_start = -1;
        while (it != cur2start.end() && it->first <= right) {
            right_bound_of_start = max(right_bound_of_start, it->second);
            values.erase(values.find(it->first - it->second));
            cur2start.erase(it++);
        }
        if (right_bound_of_start != -1 && right < w) {
            values.insert(right - right_bound_of_start);
            cur2start[right] = right_bound_of_start;
        }
        int ans = -1;
        if (values.size() > 0) {
            ans = *values.begin() + i + 1;
        }
        cout << ans << endl;
    }
}