Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she produces as much milk as possible.

Farmer John has a list of M (1 ≤ M ≤ 1,000) possibly overlapping intervals in which he is available for milking. Each interval i

 has a starting hour (0 ≤ starting_houri ≤ N), an ending hour (starting_houri < ending_houri ≤ N), and a corresponding efficiency (1 ≤ efficiencyi ≤ 1,000,000) which indicates how many gallons of milk that he can get out of Bessie in that interval. Farmer John starts and stops milking at the beginning of the starting hour and ending hour, respectively. When being milked, Bessie must be milked through an entire interval.

Even Bessie has her limitations, though. After being milked during any interval, she must rest R (1 ≤ R ≤ N) hours before she can start milking again. Given Farmer Johns list of intervals, determine the maximum amount of milk that Bessie can produce in the N hours.

Input

* Line 1: Three space-separated integers: N

, M, and R * Lines 2..M+1: Line i+1 describes FJ's ith milking interval withthree space-separated integers: starting_houri , ending_houri , and efficiencyi

Output

* Line 1: The maximum number of gallons of milk that Bessie can product in the Nhours

Sample Input

12 4 2
1 2 8
10 12 19
3 6 24
7 10 31

Sample Output

43

題意：在N時間內，有M組時間段，每個時間段對應一個價值，要求某個時間段的結束時間與另一個段的開始時間間隔大於等於R，對應價值相加，求最後能得到的最大價值的和。

思路：先按照結束時間，把時間段按升序排列，然後將第一個時間段的結束時間賦值給dp[0]，dp[i]表示從開始取到第i段所能達到的最大價值，雙重迴圈遇到第i段之前某個段結束時間加上R小於等於第i段的開始時間，則更新dp[i]。

#include<iostream>
#include<algorithm>
using namespace std;
int N,M,R;

struct hah
{
    int b;
    int e;
    int w;
};
bool cmp(hah a,hah b)
{
    if(a.e<b.e)
        return true;
    return false;
}

int main()
{

    hah m[1001];
    int dp[100001];
    cin>>N>>M>>R;
    for(int i=0; i<M; i++)
        cin>>m[i].b>>m[i].e>>m[i].w;
    sort(m,m+M,cmp);//按照結束時間從小到大排序
    int res=0;
    for( int i=0; i<M; i++)
    {
        dp[i]=m[i].w;//dp[i]表示從開始取到第i段所能達到的最大數目。
        for(int j=0; j<i; j++)//如果第i段之前某個段結束時間加上R小於等於i段的開始時間，則更新dp[i]。
            if(m[j].e+R<=m[i].b)
                dp[i]=max(dp[i],dp[j]+m[i].w);
        res=max(res,dp[i]);
    }
    cout<<res<<endl;
    return 0;
}