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>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[200010], b[200010], s1[200010], s2[200010], s3[200010], s4[200010], n, m,
    q, tn[200010];
int next[200010], last[200010], head, tt[500][500], tr[500], h[500];
int c, maxb, maxa, w, z;
long long ans[200010], x, y;
inline int qsort1(int l, int r) {
  int mid = a[s1[(l + r) >> 1]];
  int i = l;
  int j = r;
  while (true) {
    while (a[s1[i]] < mid) i++;
    while (a[s1[j]] > mid) j--;
    if (i <= j) {
      int tmp = s1[i];
      s1[i] = s1[j];
      s1[j] = tmp;
      i++;
      j--;
    }
    if (i > j) break;
  }
  if (i < r) qsort1(i, r);
  if (l < j) qsort1(l, j);
  return 0;
}
inline int qsort2(int l, int r) {
  int mid = b[s3[(l + r) >> 1]];
  int i = l;
  int j = r;
  while (true) {
    while (b[s3[i]] < mid) i++;
    while (b[s3[j]] > mid) j--;
    if (i <= j) {
      int tmp = s3[i];
      s3[i] = s3[j];
      s3[j] = tmp;
      i++;
      j--;
    }
    if (i > j) break;
  }
  if (i < r) qsort2(i, r);
  if (l < j) qsort2(l, j);
  return 0;
}
inline bool check(int i, int j, int k) {
  return (double)(ans[i] + 1ll * i * h[tn[i]] - ans[j] - 1ll * j * h[tn[j]]) /
             (i - j) <
         (double)(ans[k] + 1ll * k * h[tn[k]] - ans[j] - 1ll * j * h[tn[j]]) /
             (k - j);
}
inline void fix(int k) {
  if (k == 0) return;
  if (tn[k] > 0)
    for (int i = 1; i < tn[k]; i++) {
      h[i]++;
      while (tr[i] < tt[i][0] &&
             ans[tt[i][tr[i]]] + 1ll * tt[i][tr[i]] * h[i] <
                 ans[tt[i][tr[i] + 1]] + 1ll * tt[i][tr[i] + 1] * h[i])
        tr[i]++;
    }
  for (int i = 1; i <= q; i++)
    ans[(tn[k] - 1) * q + i] =
        ans[(tn[k] - 1) * q + i] + 1ll * ((tn[k] - 1) * q + i) * h[tn[k]];
  h[tn[k]] = 0;
  tt[tn[k]][0] = 0;
  for (int i = (tn[k] - 1) * q + 1; i <= k; i++) ans[i] = ans[i] + i;
  for (int i = 1; i <= q; i++) {
    while ((tt[tn[k]][0] >= 1 &&
            ans[tt[tn[k]][tt[tn[k]][0]]] < ans[(tn[k] - 1) * q + i]) ||
           (tt[tn[k]][0] > 1 &&
            check(tt[tn[k]][tt[tn[k]][0]], tt[tn[k]][tt[tn[k]][0] - 1],
                  (tn[k] - 1) * q + i)))
      tt[tn[k]][0]--;
    tt[tn[k]][0]++;
    tt[tn[k]][tt[tn[k]][0]] = (tn[k] - 1) * q + i;
  }
  tr[tn[k]] = 1;
  y = 0;
  z = 0;
  for (int i = 1; i <= m; i++) {
    if (ans[tt[i][tr[i]]] + 1ll * tt[i][tr[i]] * h[i] > y) {
      z = (i - 1) * q + tt[i][tr[i]];
      y = ans[tt[i][tr[i]]] + 1ll * tt[i][tr[i]] * h[i];
    }
    if (ans[tt[i][tr[i]]] + 1ll * tt[i][tr[i]] * h[i] == y && z > tt[i][tr[i]])
      z = tt[i][tr[i]];
  }
  return;
}
int main() {
  n = read();
  w = read();
  maxb = 0;
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    b[i] = read();
    maxb = max(maxb, b[i]);
    maxa = max(maxa, a[i]);
  }
  for (int i = 1; i <= n; i++) s1[i] = i;
  qsort1(1, n);
  for (int i = 1; i <= n; i++) s2[s1[i]] = i;
  for (int i = 1; i <= n; i++) s3[i] = i;
  qsort2(1, n);
  for (int i = 1; i <= n; i++) s4[s3[i]] = i;
  m = (int)sqrt(maxa) + 1;
  y = 0;
  q = maxa / m + 1;
  z = 0;
  int l = 0;
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= q; j++) tn[(i - 1) * q + j] = i;
  for (int i = 1; i <= m; i++) {
    tt[i][0] = q;
    for (int j = 1; j <= q; j++) tt[i][j] = (i - 1) * q + j;
    tr[i] = 1;
  }
  for (int i = 0; i <= maxb + 1; i++) {
    while (l < n && b[s3[l + 1]] < i) {
      l++;
      fix(a[s3[l]]);
    }
    x = 1ll * i * w * (n - l) + y;
    printf("%I64d %d\n", x, z);
  }
}
