走房间

题目网址：http://codeforces.com/problemset/problem/407/B

One day, little Vasya found himself in a maze consisting of (n + 1) rooms, numbered from 1 to (n + 1). Initially, Vasya is at the first room and to get out of the maze, he needs to get to the (n + 1)-th one.

The maze is organized as follows. Each room of the maze has two one-way portals. Let‘s consider room number i(1 ≤ i ≤ n), someone can use the first portal to move from it to room number (i + 1), also someone can use the second portal to move from it to room number p_i, where 1 ≤ p_i ≤ i.

In order not to get lost, Vasya decided to act as follows.

• Each time Vasya enters some room, he paints a cross on its ceiling. Initially, Vasya paints a cross at the ceiling of room 1.
• Let‘s assume that Vasya is in room i and has already painted a cross on its ceiling. Then, if the ceiling now contains an odd number of crosses, Vasya uses the second portal (it leads to room p_i), otherwise Vasya uses the first portal.

Help Vasya determine the number of times he needs to use portals to get to room (n + 1) in the end.

Input

The first line contains integer n (1 ≤ n ≤ 10³) — the number of rooms. The second line contains n integers p_i (1 ≤ p_i ≤ i). Each p_idenotes the number of the room, that someone can reach, if he will use the second portal in the i-th room.

Output

Print a single number — the number of portal moves the boy needs to go out of the maze. As the number can be rather large, print it modulo 1000000007(10⁹ + 7).

Sample Input

Input

21 2

Output

4

Input

41 1 2 3

Output

20

Input

51 1 1 1 1

Output

62

思路：开辟一个数组q[],q[i]表示从第i个房间的第二个门走到p[i]房间然后回到i房间的路程

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
int p[1005];
int q[1005];

int main()
{
    int n,i,j;
    long long sum,sum1;
    while(cin>>n)
    {
        sum=2;
        memset(q,0,sizeof(q));
        for(i=1;i<=n;i++)
        cin>>p[i];
        q[1]=2;
        for(i=2;i<=n;i++)
        {
            sum1=0;
            if(p[i]<i)
            {
               for(j=p[i];j<i;j++)
               {
                   sum1+=q[j];
               }
               sum1=sum1+2;
               if(sum1>=1000000007)
               sum1=sum1%1000000007;
               q[i]=sum1;
               sum+=sum1;
            }
            else
            {
                q[i]=2;
                sum=sum+2;
            }
            if(sum>=1000000007)
            sum=sum%1000000007;
        }
        cout<<sum<<endl;
    }
    return 0;
}