Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
struct Item {
  int pr, ad;
};
bool ItemCmp(const Item &a, const Item &b) { return a.ad < b.ad; }
int n, w, mxad, V, g[MAXN];
Item r[MAXN];
long long f[MAXN];
void lucky() {
  scanf("%d %d", &n, &w);
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &r[i].pr, &r[i].ad);
    V = max(V, r[i].pr), mxad = max(mxad, r[i].ad);
  }
}
namespace Div {
const int MAXB = 350;
int siz, cnt, bel[MAXN], bl[MAXB], br[MAXB];
int rec[MAXB][MAXB], qh[MAXB], qt[MAXB];
long long c[MAXN], add[MAXB];
void prep() {
  memset(bl, 0x3f, sizeof(bl));
  siz = sqrt(V), cnt = (V + siz - 1) / siz;
  for (int i = 1; i <= V; i++) {
    bel[i] = (i + siz - 1) / siz;
    bl[bel[i]] = min(bl[bel[i]], i);
    br[bel[i]] = max(br[bel[i]], i);
  }
  for (int i = 1; i <= cnt; i++) {
    qh[i] = qt[i] = 1;
    rec[i][1] = br[i];
  }
}
double slp(int i, int j) { return (double)(c[i] - c[j]) / (i - j); }
void update(int x) {
  int p = (x + siz - 1) / siz;
  for (int i = 1; i < p; i++) {
    int *Q = rec[i], &he = qh[i];
    add[i]++;
    while (he < qt[i] &&
           c[Q[he]] + add[i] * Q[he] < c[Q[he + 1]] + add[i] * Q[he + 1])
      he++;
  }
  int *st = rec[p], &top = qt[p];
  qh[p] = 1, top = 0;
  for (int i = bl[p]; i <= br[p]; i++) {
    c[i] += i * (add[p] + (i <= x));
    while (top > 1 && slp(i, st[top]) > slp(st[top], st[top - 1])) top--;
    if (!top || c[st[top]] >= c[i]) top++;
    st[top] = i;
  }
  add[p] = 0;
}
pair<long long, int> query() {
  long long mx = 0;
  int mxi = -1;
  for (int i = 1; i <= cnt; i++) {
    int p = rec[i][qh[i]];
    long long t;
    if (t = c[p] + add[i] * p, t > mx)
      mx = t, mxi = p;
    else if (t == mx && mxi == -1)
      mxi = p;
  }
  return make_pair(mx, mxi);
}
}  // namespace Div
int main() {
  lucky();
  sort(r + 1, r + n + 1, ItemCmp);
  Div::prep();
  for (int i = 1; i <= n; i++) {
    Div::update(r[i].pr);
    pair<long long, int> t = Div::query();
    f[i] = t.first, g[i] = t.second;
  }
  for (int i = 0, p = 0; i <= mxad + 1; i++) {
    while (p < n && r[p + 1].ad < i) p++;
    printf(
        "%lld"
        " %d\n",
        f[p] + (long long)(n - p) * i * w, g[p]);
  }
  return 0;
}