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() {
    int h, w; cin >> h >> w;
    vector<int> a(h), b(h);
    for (int i = 0; i < h; i++) { cin >> a[i] >> b[i]; a[i]--; b[i]--; }

    set<pair<int,int>> st;
    multiset<int> mst;
    for (int i = 0; i < w; i++) {
        st.insert(make_pair(i,i));
        mst.insert(0);
    }
    st.insert(make_pair(w,w));
    mst.insert(200200);

    for (int i = 0; i < h; i++) {
        while (true) {
            auto iter = st.lower_bound(make_pair(a[i],-1));
            if (iter->first >= b[i] + 1) break;
            int from = iter->second, to = b[i] + 1;
            if (next(iter)->first > b[i] + 1) {
                st.insert(make_pair(to, from));
                mst.insert(to - from);
            }
            mst.erase(mst.find(iter->first - iter->second));
            st.erase(iter);
        }
        int save = *mst.begin();
        cout << (save == 200200 ? -1 : save + i + 1) << endl;
    }
    return 0;
}