F. Runner's Problemtime limit per test4 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

You are running through a rectangular field. This field can be represented as a matrix with 3 rows and m columns. (i, j) denotes a cell belonging to i-th row and j-th column.

You start in (2, 1) and have to end your path in (2, m). From the cell (i, j) you may advance to:

• (i - 1, j + 1) — only if i > 1
  ,
• (i, j + 1), or
• (i + 1, j + 1) — only if i < 3.

However, there are n obstacles blocking your path. k-th obstacle is denoted by three integers ak, lk and rk, and it forbids entering any cell (ak, j) such that lk ≤ j ≤ rk.

You have to calculate the number of different paths from (2, 1) to (2, m), and print it modulo 109

 + 7.

Input

The first line contains two integers n and m (1 ≤ n ≤ 104, 3 ≤ m ≤ 1018) — the number of obstacles and the number of columns in the matrix, respectively.

Then n lines follow, each containing three integers ak, lk and rk (1 ≤ ak ≤ 3, 2 ≤ lk ≤ rk ≤ m - 1) denoting an obstacle blocking every cell (a

k, j) such that lk ≤ j ≤ rk. Some cells may be blocked by multiple obstacles.

Output

Print the number of different paths from (2, 1) to (2, m), taken modulo 109 + 7. If it is impossible to get from (2, 1) to (2, m), then the number of paths is 0.

ExampleinputCopy

2 5
1 3 4
2 2 3

output

2

思路：由於有障礙，可以將障礙的端點儲存下來，左端記錄為1，右端記錄為-1，然後進行排序。對於每個端點，如果是1，則表示當前行有障礙，記錄在sum陣列中，即sum[i]++；如果是-1，sum[i]--。然後進行快速冪的時候，如果sum[i]不等於0，就說明該行有障礙，友矩陣就需要改變。由於是一段一段的轉移，要用一個數記錄上一次轉移到了什麼位置，每次轉移的時候，從上一個位置轉移到當前端點。

#include<bits/stdc++.h>
using namespace std;
const int MAX=1e6;
const int MOD=1e9+7;
const double PI=acos(-1);
typedef __int64 ll;
int sum[4];
ll a[3];//分別儲存每行的方案數
struct data{ll b[3][3];};
struct lenka
{
    ll x,y,in;
}A[MAX];
int cmp(const lenka& p,const lenka& q){return p.y<q.y;}//對端點排序
data cla(const data& p,const data& q)
{
    data c;
    memset(c.b,0,sizeof c.b);
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {
            for(int k=0;k<3;k++)
            {
                c.b[i][j]+=p.b[i][k]*q.b[k][j]%MOD;
                c.b[i][j]%=MOD;
            }
        }
    }
    return c;
}
void POW(ll n)
{
    data res,b;
    memset(res.b,0,sizeof res.b);
    memset(b.b,0,sizeof b.b);
    for(int i=0;i<3;i++)res.b[i][i]=1;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)b.b[i][j]=(sum[j]==0);//根據當前行是否有障礙來確定友矩陣
    }
    b.b[0][2]=b.b[2][0]=0;
    while(n)
    {
        if(n&1)res=cla(b,res);
        b=cla(b,b);
        n/=2;
    }
    ll x=(a[0]*res.b[0][0]%MOD+a[1]*res.b[1][0]%MOD+a[2]*res.b[2][0]%MOD)%MOD;
    ll y=(a[0]*res.b[0][1]%MOD+a[1]*res.b[1][1]%MOD+a[2]*res.b[2][1]%MOD)%MOD;
    ll z=(a[0]*res.b[0][2]%MOD+a[1]*res.b[1][2]%MOD+a[2]*res.b[2][2]%MOD)%MOD;
    a[0]=x;
    a[1]=y;
    a[2]=z;//更新方案數
}
int main()
{
    ll n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        ll x,L,R;
        scanf("%I64d%I64d%I64d",&x,&L,&R);
        A[i].x=x-1;
        A[i].y=L-1;
        A[i].in=1;
        A[i+n].x=x-1;
        A[i+n].y=R;
        A[i+n].in=-1;
    }
    sort(A+1,A+2*n+1,cmp);
    memset(sum,0,sizeof sum);
    memset(a,0,sizeof a);
    a[1]=1;
    ll L=1;//記錄一次更新到的位置
    for(int i=1;i<=2*n;i++)
    {
        POW(A[i].y-L);
        L=A[i].y;
        sum[A[i].x]+=A[i].in;
    }
    POW(m-L);
    cout<<a[1]%MOD<<endl;
    return 0;
}