output

standard output

There are nn TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The ii-th of the shows is going from lili-th to riri-th minute, both ends inclusive.

You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [li,ri][li,ri] and [lj,rj][lj,rj] intersect, then shows ii and jj can't be watched simultaneously on one TV.

Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.

There is a TV Rental shop near you. It rents a TV for xx rupees, and charges yy (y<xy<x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a;b][a;b] you will need to pay x+y⋅(b−a)x+y⋅(b−a).

You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 109+7109+7.

Input

The first line contains integers nn, xx and yy (1≤n≤1051≤n≤105, 1≤y<x≤1091≤y<x≤109) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.

Each of the next nn lines contains two integers lili and riri (1≤li≤ri≤1091≤li≤ri≤109) denoting the start and the end minute of the ii-th TV show.

Output

Print exactly one integer — the minimum cost to view all the shows taken modulo 109+7109+7.

Examples

input

5 4 3
1 2
4 10
2 4
10 11
5 9

output

60

input

6 3 2
8 20
6 22
4 15
20 28
17 25
20 27

output

142

input

2 1000000000 2
1 2
2 3

output

999999997

Note

In the first example, the optimal strategy would be to rent 33 TVs to watch:

• Show [1,2][1,2] on the first TV,
• Show [4,10][4,10] on the second TV,
• Shows [2,4],[5,9],[10,11][2,4],[5,9],[10,11] on the third TV.

This way the cost for the first TV is 4+3⋅(2−1)=74+3⋅(2−1)=7, for the second is 4+3⋅(10−4)=224+3⋅(10−4)=22 and for the third is 4+3⋅(11−2)=314+3⋅(11−2)=31, which gives 6060 int total.

In the second example, it is optimal watch each show on a new TV.

In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 109+7109+7.

思路：首先排序是肯定可以想到的，但關鍵是用什麼來維護比較你一個節目接在另一個下面的費用小還是直接在開一臺電視的費用小。大佬們又multiset，也有用set加map來維護的。我就使用set加map維護的（注意比較的時候有個二分找結束的時間用set本身的二分函式效率快很多）。用到二分查詢的原因是我們記錄結束的時間那麼新的時間加入就要找離開始時間最近的結束時間。複雜度是nlogn

程式碼：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#define LL long long

using namespace std;
const int maxn=1e5+100;
const int mod=1e9+7;
struct node
{
    LL a,b;
    friend bool operator <(node p,node q)
    {
        if(p.a==q.a)
        {
            return p.b<q.b;
        }
        return p.a<q.a;
    }
};
node c[maxn];
set<LL>st;
map<LL,int>mp;
int main()
{
    int n;
    LL x,y;
    scanf("%d%lld%lld",&n,&x,&y);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&c[i].a,&c[i].b);
    }
    sort(c+1,c+1+n);
    LL ans=0;
    st.insert(c[1].b);
    mp[c[1].b]++;
    ans=(ans+x+(c[1].b-c[1].a)*y%mod)%mod;
    set<LL>::iterator it;
    for(int i=2;i<=n;i++)
    {
        it=st.lower_bound(c[i].a);

        if(it==st.begin())
        {
            ans=(ans+x+(c[i].b-c[i].a)*y%mod)%mod;
            mp[c[i].b]++;
        }
        else
        {
            it--;
            LL value=*it;
            if(c[i].a>value&&(c[i].a-value)*y<=x)
            {
                mp[value]--;
                if(mp[value]==0)
                    st.erase(value);
                ans=(ans+(c[i].b-value)*y%mod)%mod;
                mp[c[i].b]++;
            }
            else
            {
                ans=(ans+x+(c[i].b-c[i].a)*y%mod)%mod;
                mp[c[i].b]++;
            }
        }
        st.insert(c[i].b);
    }
    printf("%lld\n",ans);
    return 0;

}