Codeforces Round #252(Div. 2) 441B. Valera and Fruits 贪心

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 b_i fruit grow, they will ripen on a day number a_i. Unfortunately, the fruit on the tree get withered, so they can only be collected on day a_i and day a_i + 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 a_i and b_i(1 ≤ a_i, b_i ≤ 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.

Examples

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.

题意：

n棵果树，第i棵的产量为bi，收获时间为ai和ai+1两天，过期不能收获，问最多能收获多少果实。

题解：

当然先收获早成熟的，所以我们按a升序排序，以此往后收。

注意，如果两个果树a相等，就把两棵树合并，因为果实是没有区别的，所以合并处理就好了。

/****************
*PID:441b div2
*Auth:Jonariguez
*****************
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

const int maxn=3000+10;
int vis[maxn];
struct pp{
    int a,b;
}t[maxn],s[maxn];

int cmp(const pp &x,const pp &y){
    if(x.a==y.a)
        return x.b>y.b;
    return x.a<y.a;
}

int main()
{
    int i,j,n,v;
    while(scanf("%d%d",&n,&v)!=EOF){
        for(i=1;i<=n;++i){
            sc(t[i].a);sc(t[i].b);
        }
        sort(t+1,t+n+1,cmp);
        s[1]=t[1];
        int last=1,tot=1;
        for(i=2;i<=n;i++){
            if(t[i].a==s[last].a){
                s[last].b+=t[i].b;
            }else {
                s[++tot]=t[i];
                last=tot;
            }
        }
        int res=0,now=1;
        last=0;
        j=1;s[tot+1].a=3333;
        for(now=1;now<=3001;now++){
            if(now==s[j].a){
                int temp=0;
                if(last){
                    if(v>=last){
                        res+=last;
                        temp=v-last;
                        if(temp<=s[j].b){
                            res+=temp;last=s[j].b-temp;
                        }else {
                            res+=s[j].b;last=0;
                        }
                    }else {
                        res+=v;last=s[j].b;
                    }
                }else {
                    res+=min(v,s[j].b);
                    last=max(0,s[j].b-v);
                }
                j++;
            }else {
                res+=min(v,last);last=0;
            }
        }
        pfn(res);
    }
    return 0;
}