#include <iostream>
#include <math.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;     //无穷大   
int a[1005];
int b[1005];
int dp[1005][1005][2];       
int sum[1005];          //存放第1到i人的怒气值

struct node
{
    int x,b;        
} s[1005];

bool cmp(node x,node y)      //按照坐标从大到小排序
{
    return x.x<y.x;
}
int main()
{
    int n,v,x,i,j,l;
    while(scanf("%d%d%d",&n,&v,&x)!=EOF)
    {
        for(i=1;i<=n;i++)
            scanf("%d%d",&s[i].x,&s[i].b);    
        memset(dp,INF,sizeof(dp));     //方便后面判断        
        sort(s+1,s+n+1,cmp);
        for(i=1;i<=n;i++)
            sum[i]=sum[i-1]+s[i].b;
        for(i=1;i<=n;i++)              //因为要怒气最小送完一个区间只可能在区间端点，dp[i][j][0]表示送完[i，j]区间后在区间左边，反之在右边
            dp[i][i][0]=dp[i][i][1]=abs(s[i].x-x)*sum[n];
        for(l=2;l<=n;l++)                //长度为1的区间已被赋初值，从2到n进行递归
            for(i=1;i<=n-l+1;i++)
            {
                j=i+l-1;          //区间dp真的巧妙
                //对每个长度大于等于2的区间[i,j]，都是由4种长度小1的区间变化而来
                dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][0]+abs(s[i+1].x-s[i].x)*(sum[n]-sum[j]+sum[i]));  //上一步时[i+1,j]均已被送达，不产生怒气，下一秒产生的总怒气值为sum[n]-sum[j]+sum[i](类似数轴）
                dp[i][j][0]=min(dp[i][j][0],dp[i+1][j][1]+abs(s[j].x-s[i].x)*(sum[n]-sum[j]+sum[i]));
                dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][0]+abs(s[j].x-s[i].x)*(sum[n]-sum[j-1]+sum[i-1]));
                dp[i][j][1]=min(dp[i][j][1],dp[i][j-1][1]+abs(s[j].x-s[j-1].x)*(sum[n]-sum[j-1]+sum[i-1]));
            }
            printf("%d\n",min(dp[1][n][0],dp[1][n][1])*v);  //速度为1/v，ans/(1/v)=ans*v)
    }
    return 0;
}

When we are focusing on solving problems, we usually prefer to stay in front of computers rather than go out for lunch. At this time, we may call for food delivery.

Suppose there are N people living in a straight street that is just lies on an X-coordinate axis. The ith person's coordinate is Xi meters. And in the street there is a take-out restaurant which has coordinates X meters. One day at lunchtime, each person takes an order from the restaurant at the same time. As a worker in the restaurant, you need to start from the restaurant, send food to the N people, and then come back to the restaurant. Your speed is V-1 meters per minute.

You know that the N people have different personal characters; therefore they have different feeling on the time their food arrives. Their feelings are measured by Displeasure Index. At the beginning, the Displeasure Index for each person is 0. When waiting for the food, the ith person will gain Bi Displeasure Index per minute.

If one's Displeasure Index goes too high, he will not buy your food any more. So you need to keep the sum of all people's Displeasure Index as low as possible in order to maximize your income. Your task is to find the minimal sum of Displeasure Index.

Input

The input contains multiple test cases, separated with a blank line. Each case is started with three integers N ( 1 <= N <= 1000 ), V ( V > 0), X ( X >= 0 ), then N lines followed. Each line contains two integers Xi ( Xi >= 0 ), Bi ( Bi >= 0), which are described above.

You can safely assume that all numbers in the input and output will be less than 231 - 1.

Please process to the end-of-file.

Output

For each test case please output a single number, which is the minimal sum of Displeasure Index. One test case per line.

Sample Input

5 1 0
1 1
2 2
3 3
4 4
5 5
 

Sample Output

55