Valera and Fruits

B. Valera and Fruits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Valera loves his garden, where n fruit trees grow.

This year he will enjoy a great harvest! On the i-th tree bi fruit grow, they will ripen on a day number ai. Unfortunately, the fruit on the tree get withered, so they can only be collected on day ai and day ai + 1 (all fruits that are not collected in these two days, become unfit to eat).

Valera is not very fast, but there are some positive points. Valera is ready to work every day. In one day, Valera can collect no more thanv fruits. The fruits may be either from the same tree, or from different ones. What is the maximum amount of fruit Valera can collect for all time, if he operates optimally well?

Input

The first line contains two space-separated integers n and v (1 ≤ n, v ≤ 3000) — the number of fruit trees in the garden and the number of fruits that Valera can collect in a day.

Next n lines contain the description of trees in the garden. The i-th line contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ 3000) — the day the fruits ripen on the i-th tree and the number of fruits on the i-th tree.

Output

Print a single integer — the maximum number of fruit that Valera can collect.

Sample test(s)

input

2 3
1 5
2 3

output

8

input

5 10
3 20
2 20
1 20
4 20
5 20

output

60

Note

In the first sample, in order to obtain the optimal answer, you should act as follows.

• On the first day collect 3 fruits from the 1-st tree.
• On the second day collect 1 fruit from the 2-nd tree and 2 fruits from the 1-st tree.
• On the third day collect the remaining fruits from the 2-nd tree.

In the second sample, you can only collect 60 fruits, the remaining fruit will simply wither.

题目比較简单，但有两个easy忽略的地方，代码中已经标注了。

#include <iostream>
#include <cstdio>
using namespace std;
int t[3333];
int main()
{
    int n,v,m,i,j,a,b;
    cin>>n>>v;
    int minn=9999,maxx=-1;
    for(i=0;i<n;i++)
    {
        scanf("%d %d",&a,&b);
        t[a]+=b;//可能多棵树在同一天成熟
        maxx=max(maxx,a);
        minn=min(minn,a);
    }
    int sum=0,tmp=0;

    for(i=minn;i<=maxx+1;i++)
    {
        if(tmp>=v)
        {
            tmp=t[i];
            sum+=v;
        }
        else
        {
            if(tmp+t[i]<=v)
            {


                sum+=tmp+t[i];
                tmp=0;//这里tmp位置非常重要，之前放在sum前面，就悲剧了
            }
            else
            {
                tmp=t[i]-(v-tmp);
                sum+=v;
            }
        }
      //  cout<<tmp<<" "<<sum<<endl;
    }
    cout<<sum<<endl;
    return 0;
}