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;
const int N = 2e5 + 5;
int l[N], r[N], n , m;
map <int, int> mp;
int cnt[N];
inline int read () {
    int tot = 0, f = 1; char c = getchar ();
    while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar (); }
    while (c >= '0' && c <= '9') { tot = tot * 10 + c - '0'; c = getchar (); }
    return tot * f;
}
signed main () {
    n = read (); m = read ();
    for (int i = 1; i <= n; i++) {
        l[i] = read (); r[i] = read ();
    }
    for (int i = 0; i < m; i++) mp[i] = i;
    cnt[0] = m;
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        l[i]--; r[i]--;
        auto it = mp.lower_bound (l[i]);
        int pos = -1;
        while (it != mp.end () && it->first <= r[i] + 1) {
            pos = max (pos, it->second);
            cnt[it->first - it->second] --;
            it = mp.erase (it);
        }
        if (pos >= 0 && r[i] + 1 < m) {
            mp[r[i] + 1] = pos;
            cnt[r[i] + 1 - pos]++;
        }
        while (!cnt[ans] && ans < m) ans++;
        if (ans == m) printf ("-1\n");
        else printf ("%d\n", i + ans);
    }
    return 0;
}