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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using pdd = pair<double, double>;
using uint = unsigned int;
template <typename T>
using Graph = vector<vector<T>>;
const int MOD = 1e9 + 7;
const ld PI = acosl(-1);

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int H, W;
    cin >> H >> W;
    set<pii> st;
    multiset<int> mst;
    for (int i = 0; i < W; ++i) {
        st.emplace((pii){i, i});
        mst.emplace(0);
    }

    for (int i = 0; i < H; ++i) {
        int A, B;
        cin >> A >> B;
        A--, B--;

        auto itr = st.lower_bound({A, 0});
        int maidx = -1;
        while (itr != st.end() && itr->first <= B) {
            mst.erase(mst.find(itr->first - itr->second));
            maidx = itr->second;
            auto citr = itr;
            ++itr;
            st.erase(citr);
        }

        if (B + 1 < W && ~maidx) {
            st.emplace((pii){B + 1, maidx});
            mst.emplace(B + 1 - maidx);
        }

        if (mst.empty())
            cout << -1 << '\n';
        else
            cout << *mst.begin() + i + 1 << '\n';
    }

    return 0;
}