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;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const int N = 100005;

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    set<pair<int, int> > se;
    multiset<int> ans;
    for (int i = 1; i <= m; i++) {
        se.insert({i, i});
        ans.insert({0});
    }
    for (int i = 1; i <= n; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        if (y == m) y = INF;
        auto it = se.lower_bound({x, 0});
        int t = -1;
        for (; it != se.end() && it->first <= y;) {
            ans.erase(ans.find(it->first - it->second));
            t = max(t, it->second);
            it = se.erase(it);
        }
        if (t > -1) {
            ans.insert(y + 1 - t);
            se.insert({y + 1, t});
        }
        int r = *ans.begin();
        if (r > m)
            puts("-1");
        else
            printf("%d\n", r + i);
    }
    return 0;
}