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;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define ALL(v) (v).begin(), (v).end()
using ll = long long;
constexpr int INF = 1e9;
constexpr long long LINF = 1e18;
constexpr long long MOD = 1e9 + 7;

signed main() {
    int h, w;
    cin >> h >> w;
    int a[h], b[h];
    rep(i, h) {
        cin >> a[i] >> b[i];
        a[i]--;
        b[i]--;
    }
    set<pair<int, int>> S;  // {現在の位置, 始めの位置}
    multiset<int> M;        // 現在の位置 - 始めの位置
    for (int i = 0; i < w; i++) {
        S.emplace(i, i);
        M.emplace(0);
    }
    rep(i, h) {
        int mx = -1;
        auto itr = S.lower_bound({a[i], 0});
        while (itr != S.end() &&
               itr->first <= b[i]) {  // 現在の位置が範囲内にある
            mx = max(mx, itr->second);
            M.erase(M.find(itr->first - itr->second));
            itr = S.erase(itr);  // set::eraseの返り値は次のイテレータ
        }
        if (mx != -1 &&
            b[i] + 1 <
                w) {  // 始めの位置が最も右にある点だけ残す(ただし、範囲外にでるときは残さない)
            S.emplace(b[i] + 1, mx);
            M.emplace(b[i] + 1 - mx);
        }
        if (M.empty()) {
            cout << -1 << endl;
            continue;
        }
        cout << *M.begin() + i + 1 << endl;
    }
    return 0;
}