首先考慮怎麼安排攻擊順序。顯然如果攻擊了某臺兵器就應該一直連續攻擊直到將其破壞，破壞所需時間可以直接算出來，設其為b。假設確定了某個破壞順序，如果交換相鄰兩個兵器，顯然不會對其他兵器造成影響，兩種順序的代價則分別為a₁(b₁-1)+a₂(b₁+b₂-1)和a₁(b₁+b₂-1)+a₂(b₂-1)，那麼當a₂b₁<a₁b₂時先破壞1較優。於是按b/a從小到大排序。

　　然後考慮怎麼秒殺。如果只能秒殺一個顯然直接列舉即可。假設已確定要秒殺的是第i個，則需要找到j>i最小化Σa_x(B_x-1)-a_i(B_i-1)-a_j(B_j-1)-(A_n-A_i)b_i-(A_n-A_j)b_j+a_jb_i 

#include<iostream> 
#include
 
<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 300010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return
 
 c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,q[N];
ll ans,tot,c[N];
struct data{int x,y;ll z;
}a[N],b[N];
bool cmp(const data&a,const data&b)
{
    return b.x*a.y<a.x*b.y;
}
bool cmp2(const data&a,const data&b) 
{
    return a.y>b.y;
}
bool cmp3(const data&a,const data&b)
{
    return a.x<b.x;
}
ll calc(int x,int y)
{
    return tot-a[x].z-a[y].z+a[y].x*a[x].y;
}
double slope(int i,int j)
{
    return (double)(a[j].z-a[i].z)/(a[j].x-a[i].x);
}
void solve(int l,int r)
{
    if (l>=r) return;
    int mid=l+r>>1;
    solve(l,mid);
    solve(mid+1,r);
    sort(a+l,a+mid+1,cmp2);
    sort(a+mid+1,a+r+1,cmp3);
    int head=0,tail=0;
    for (int i=mid+1;i<=r;i++)
    {
        while (head<tail&&slope(q[tail-1],q[tail])<slope(q[tail],i)) tail--;
        q[++tail]=i;
    }
    for (int i=l;i<=mid;i++)
    {
        while (head<tail&&slope(q[head],q[head+1])>a[i].y) head++;
        ans=min(ans,calc(i,q[head]));
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4700.in","r",stdin);
    freopen("bzoj4700.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    ll A=0,B=0;
    for (int i=1;i<=n;i++) (A+=a[i].x=read()),a[i].y=(read()-1)/m+1;
    sort(a+1,a+n+1,cmp);
    for (int i=1;i<=n;i++)
    {
        A-=a[i].x,B+=a[i].y;
        a[i].z=a[i].x*(B-1)+A*a[i].y;
        tot+=a[i].x*(B-1);
    }
    ans=tot;
    solve(1,n);
    cout<<ans;
    return 0;
}