All modern mobile applications are divided into free and paid. Even a single application developers often release two versions: a paid version without ads and a free version with ads.

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Suppose that a paid version of the app costs p (p is an integer) rubles, and the free version of the application contains c ad banners. Each user can be described by two integers: ai — the number of rubles this user is willing to pay for the paid version of the application, and bi — the number of banners he is willing to tolerate in the free version.

The behavior of each member shall be considered strictly deterministic:

  * if for user i, value bi is at least c, then he uses the free version, 
  * otherwise, if value ai is at least p, then he buys the paid version without advertising, 
  * otherwise the user simply does not use the application. 



Each user of the free version brings the profit of c × w rubles. Each user of the paid version brings the profit of p rubles.

Your task is to help the application developers to select the optimal parameters p and c. Namely, knowing all the characteristics of users, for each value of c from 0 to (max bi) + 1 you need to determine the maximum profit from the application and the corresponding parameter p.

Input

The first line contains two integers n and w (1 ≤ n ≤ 105; 1 ≤ w ≤ 105) — the number of users and the profit from a single banner. Each of the next n lines contains two integers ai and bi (0 ≤ ai, bi ≤ 105) — the characteristics of the i-th user.

Output

Print (max bi) + 2 lines, in the i-th line print two integers: pay — the maximum gained profit at c = i - 1, p (0 ≤ p ≤ 109) — the corresponding optimal app cost. If there are multiple optimal solutions, print any of them.

Examples

Input

2 1
2 0
0 2


Output

0 3
3 2
4 2
2 2


Input

3 1
3 1
2 2
1 3


Output

0 4
3 4
7 3
7 2
4 2
#include <bits/stdc++.h>
std::vector<int> v[111111];
int n, ctb, w, a[111111], b[111111];
const int blk = 315;
void emax(int& x, int y) { x < y ? x = y : 0; }
inline int bel(int x) { return (x - 1) / blk + 1; }
inline int Lol(int x) { return (x - 1) * blk + 1; }
inline int Rol(int x) { return x * blk; }
struct Block {
  long long Dat[444], Max, MxId;
  int crash, delt, L;
  void rebuild() {
    Max = Dat[1], MxId = L, delt = 0, crash = 1e9;
    register int i, t = 1;
    for (i = 2; i <= blk; i++)
      if (Dat[i] >= Max) Max = Dat[i], MxId = L + i - 1, t = i;
    for (i = t + 1; i <= blk; i++) {
      long long d = i - t, dv = Max - Dat[i], rc = (dv - 1) / d + 1;
      if (crash > rc) crash = rc;
    }
  }
  void run() {
    delt++;
    if (delt == crash) {
      for (register int i = 1; i <= blk; i++)
        Dat[i] += delt * 1ll * (L + i - 1);
      rebuild();
    } else
      Max += MxId;
  }
  void push(int rd) {
    for (register int i = 1; i <= blk; i++)
      Dat[i] += (delt + (i <= rd)) * 1ll * (L + i - 1);
    rebuild();
  }
} B[444];
void ins(int x) {
  register int i;
  for (i = 1; i < bel(x); i++) B[i].run();
  B[i].push(x - (i - 1) * blk);
}
void qmax(long long delt) {
  long long mxv = 0;
  int mxp = 0;
  register int i;
  for (i = 1; i <= ctb; i++)
    if (B[i].Max > mxv) mxv = B[i].Max, mxp = B[i].MxId;
  printf("%lld %d\n", delt + mxv, mxp);
}
int main() {
  scanf("%d%d", &n, &w);
  register int i, ma = 0, mb = 0;
  for (i = 1; i <= n; i++)
    scanf("%d%d", a + i, b + i), v[b[i]].push_back(a[i]), emax(mb, b[i]),
        emax(ma, a[i]);
  for (ctb = bel(ma), i = 1; i <= ctb; i++) B[i].L = Lol(i), B[i].rebuild();
  int cp = n;
  for (i = 0; i <= mb + 1; i++) {
    if (!i)
      puts("0 998244353");
    else
      qmax(cp * 1ll * i * w);
    for (int t : v[i]) ins(t), cp--;
  }
}
