CodeForces 407B

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

const int mod = 1000000000 + 7;#include<stdio.h>#include<iostream>using namespace std;int main(){    long long dp[1111];    int a[1111],n,i;    cin>>n;    for(i=1;i<=n;i++)    {        cin>>a[i];    }    dp[1]=0;    for(i=1;i<=n;i++)    {        dp[i+1]=(2*dp[i]-dp[a[i]]+2+mod)%mod;    }    cout<<dp[n+1]<<endl;    return 0;}

Description

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 numberi(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 numberp_i, where1 ≤ 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 room1.
• 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 roomp_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 containsn integers p_i (1 ≤ p_i ≤ i). Eachp_i denotes the number of the room, that someone can reach, if he will use the second portal in thei-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 modulo1000000007(10⁹ + 7).

Sample Input

Input

21 2

Output

4

Input

41 1 2 3

Output

20

Input

51 1 1 1 1

Output

62