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 <iostream>

using namespace std;

const int kMaxN = 200001;

struct V {
  int c, v;
} v[kMaxN * 4];
int w, h, x, y, z;

void P(int i, int ml, int c) {
  v[i].c = c;
  v[i].v = c != -1 ? ml - c : w;
}

void U(int i, int ml, int mr, int nl, int nr, int c) {
  if (nl == ml && nr == mr) {
    P(i, ml, c);
    return;
  }
  int m = (ml + mr) >> 1;
  if (v[i].c) {
    P(i * 2, ml, v[i].c);
    P(i * 2 + 1, m + 1, v[i].c);
    v[i].c = 0;
  }
  if (nl <= m) {
    U(i * 2, ml, m, nl, min(nr, m), c);
  }
  if (nr > m) {
    U(i * 2 + 1, m + 1, mr, max(m + 1, nl), nr, c);
  }
  v[i].v = min(v[i * 2].v, v[i * 2 + 1].v);
}

void F(int i, int ml, int mr, int t) {
  if (v[i].c) {
    z = v[i].c;
    return;
  }
  int m = (ml + mr) >> 1;
  if (t <= m) {
    F(i * 2, ml, m, t);
  } else {
    F(i * 2 + 1, m + 1, mr, t);
  }
}

int main() {
  cin >> h >> w;
  for (int i = 1; i <= w; i++) {
    U(1, 1, w, i, i, i);
  }
  for (int i = 1; i <= h; i++) {
    cin >> x >> y;
    if (x == 1) {
      U(1, 1, w, x, y, -1);
    } else {
      F(1, 1, w, x - 1);
      U(1, 1, w, x, y, z);
    }
    cout << (v[1].v == w ? -1 : v[1].v + i) << endl;
  }
  return 0;
}