//看看会不会爆int!数组会不会少了一维！

//取物问题一定要小心先手胜利的条件

#include <bits/stdc++.h>

using namespace std;

#define LL long long

#define ALL(a) a.begin(), a.end()

#define pb push_back

#define mk make_pair

#define fi first

#define se second

#define haha printf("haha\n")

const int maxn = 1e5 + ;

int a[maxn];

int n;

struct Tree{

    int lpos, rpos;

    Tree(int l = , int r = ): lpos(l), rpos(r){}

}tree[maxn << ];

inline void push_up(int o){

    int lb = o << , rb = o <<  | ;

    tree[o].rpos = max(tree[lb].rpos, tree[rb].rpos);

    if (tree[lb].rpos >= tree[rb].rpos) {

        tree[o].lpos = tree[lb].lpos;

    }

    else tree[o].lpos = tree[rb].lpos;

}

void build_tree(int l, int r, int o){

    if (l == r){

        tree[o].rpos = a[l];

        tree[o].lpos = l;

        return ;

    }

    int mid = (l + r) / ;

    if (l <= mid) build_tree(l, mid, o << );

    if (r > mid) build_tree(mid + , r, o <<  | );

    push_up(o);

}

Tree query(int l ,int r, int ql, int qr, int o){

    if (ql <= l && qr >= r){

        return tree[o];

    }

    Tree ans, re;

    ans.lpos = re.lpos = ans.rpos = re.rpos = ;

    int mid = (l + r) / ;

    if (ql <= mid)

        ans = query(l, mid, ql, qr, o << );

    if (qr > mid)

        re = query(mid + , r, ql, qr, o <<  | );

    if (ans.rpos < re.rpos) ans = re;

    return ans;

}

LL dp[maxn];///定义从i开始走到n所需要的最小花费

int main(){

    scanf("%d", &n);

    a[n] = n;

    for (int i = ; i <= n - ; i++) scanf("%d", a + i);

    memset(tree, , sizeof(tree));

    build_tree(, n, );

    LL ans = ;

    for (int i = n - ; i > ; i--){

        Tree pos = query(, n, i + , a[i], );

        dp[i] = n - i + dp[pos.lpos] - (a[i] - pos.lpos);

        ans += 1LL * dp[i];

    }

    printf("%lld\n", ans);

    return ;

}