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 n, k; scanf("%d %d", &k, &n);
    map<int, int> m;
    multiset<int> s;
    for (int i = 1; i <= n; ++i)
        m[i] = 0, s.insert(0);
    for (int i = 1; i <= k; ++i) {
        int l, r; scanf("%d %d", &l, &r);
        int x = 0, y = n + 1;
        auto it = m.lower_bound(l);
        while (it != m.end() && it->first <= r + 1) {
            y = min(y, r + 1 - it->first + it->second);
            s.erase(s.find(it->second));
            it = m.erase(it);
        }
        if (y != n + 1 && r < n) {
            m[r + 1] = y;
            s.insert(y);
        }
        printf("%d\n", s.empty() ? -1 : i + *s.begin());
    }
    return 0;
}