Codeforces Round #353 (Div. 2) E. Trains and Statistic

Vasya commutes by train every day. There are n train stations in the city, and at thei-th station it's possible to buy only tickets to stations fromi + 1 to a_i inclusive. No tickets are sold at the last station.

Let ρ_i, j be the minimum number of tickets one needs to buy in order to get from stationsi to station j. As Vasya is fond of different useless statistic he asks you to compute the sum of all valuesρ_i, j among all pairs1 ≤ i < j ≤ n.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of stations.

The second line contains n - 1 integer a_i (i + 1 ≤ a_i ≤ n), thei-th of them means that at the i-th station one may buy tickets to each station from i + 1 to a_i inclusive.

Output

Print the sum of ρ_i, j among all pairs of1 ≤ i < j ≤ n.

Examples

Input

44 4 4

Output

6

Input

52 3 5 5

Output

17

Note

In the first sample it's possible to get from any station to any other (with greater index) using only one ticket. The total number of pairs is6, so the answer is also 6.

Consider the second sample:

• ρ_1, 2 = 1
• ρ_1, 3 = 2
• ρ_1, 4 = 3
• ρ_1, 5 = 3
• ρ_2, 3 = 1
• ρ_2, 4 = 2
• ρ_2, 5 = 2
• ρ_3, 4 = 1
• ρ_3, 5 = 1
• ρ_4, 5 = 1

Thus the answer equals 1 + 2 + 3 + 3 + 1 + 2 + 2 + 1 + 1 + 1 = 17.

题意：有n个车站，每个车站有一个可达区间[i+1,ai]，问所有车站的最小换站距离之和。

分析：设i为起点，k为[i+1,a[i]]中a[k]最大的车站编号，则对于任意j>ai，选择从k换站一定是最好的。

#include <cmath>#include <cstdio>#include <iostream>#include <algorithm>#define got(x) (1<<x) using namespace std;long long ans,dp[100007];int n,a[100007],Max[100007][28];int gotmax(int x,int y){int l = log2(y-x+1);return a[Max[x][l]] < a[Max[y-got(l)+1][l]] ? Max[y-got(l)+1][l] : Max[x][l];}int main(){scanf("%d",&n);for(int i = 1;i < n;i++){scanf("%d",&a[i]);Max[i][0] = i;}for(int i = 1;got(i) < n;i++) for(int j = 1;j+got(i)-1 < n;j++)  Max[j][i] = a[Max[j][i-1]] < a[Max[j+got(i-1)][i-1]] ? Max[j+got(i-1)][i-1]:Max[j][i-1];dp[n] = 0;for(int i = n-1;i;i--) {if(a[i] == n){dp[i] = 1ll*(n - i);ans += dp[i];continue;}int k = gotmax(i+1,a[i]);dp[i] = dp[k] + 1ll*(n - i - a[i] + k);ans += dp[i];}cout<<ans<<endl;}