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 N = 100005, B = 305;
int n, W, cnt, mx;
int tg[N], q[N];
int pos[N], now[N];
int L[N], R[N];
long long sum[N];
pair<long long, long long> res[N], ans;
vector<int> vec[N];
bool cmp(int x, int y, int z) {
  return 1ll * (z - x) * (sum[y] - sum[x]) <= 1ll * (y - x) * (sum[z] - sum[x]);
}
long long F(int p, int dif) { return 1ll * p * dif + sum[p]; }
void build(int x) {
  pos[x] = L[x] - 1;
  now[x] = L[x];
  for (int i = (int)(L[x]); i <= (int)(R[x]); i++) {
    for (; pos[x] > L[x] && cmp(q[pos[x] - 1], q[pos[x]], i); --pos[x])
      ;
    q[++pos[x]] = i;
  }
}
void UPD(int x) {
  for (; now[x] < pos[x] && F(q[now[x]], tg[x]) < F(q[now[x] + 1], tg[x]);
       ++now[x])
    ;
}
pair<long long, long long> query(int x) {
  return pair<long long, long long>(F(q[now[x]], tg[x]), q[now[x]]);
}
void build_blk() {
  cnt = (100000 - 1) / B + 1;
  for (int i = (int)(1); i <= (int)(cnt); i++) {
    L[i] = (i - 1) * B + 1;
    R[i] = min(i * B, 100000);
    build(i);
    UPD(i);
  }
}
void change(int x) {
  int be = (x - 1) / B + 1;
  for (int i = (int)(1); i <= (int)(be - 1); i++) ++tg[i], UPD(i);
  for (int i = (int)(L[be]); i <= (int)(x); i++) sum[i] += i;
  build(be);
  UPD(be);
}
int main() {
  scanf("%d%d", &n, &W);
  for (int i = (int)(1); i <= (int)(n); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    mx = max(mx, y);
    vec[y].push_back(x);
  }
  build_blk();
  int sum = n;
  for (int i = (int)(0); i <= (int)(mx + 1); i++) {
    printf("%lld %lld\n", ans.first + 1ll * i * W * sum, ans.second);
    for (auto j : vec[i]) change(j), sum--;
    for (int j = (int)(1); j <= (int)(cnt); j++) ans = max(ans, query(j));
  }
}