题干：

Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city.

City consists of n intersections numbered from 1 to n. Mike starts walking from his house located at the intersection number 1 and goes along some sequence of intersections. Walking from intersection number i to intersection j requires |i - j|units of energy. The total energy spent by Mike to visit a sequence of intersections p1 = 1, p2, ..., pk is equal to  units of energy.

Of course, walking would be boring if there were no shortcuts. A shortcut is a special path that allows Mike walking from one intersection to another requiring only 1 unit of energy. There are exactly n shortcuts in Mike's city, the ith of them allows walking from intersection i to intersection ai (i ≤ ai ≤ ai + 1) (but not in the opposite direction), thus there is exactly one shortcut starting at each intersection. Formally, if Mike chooses a sequence p1 = 1, p2, ..., pk then for each 1 ≤ i < k satisfying pi + 1 = api and api ≠ pi Mike will spend only 1 unit of energyinstead of |pi - pi + 1| walking from the intersection pi to intersection pi + 1. For example, if Mike chooses a sequence p1 = 1, p2 = ap1, p3 = ap2, ..., pk = apk - 1, he spends exactly k - 1 units of total energy walking around them.

Before going on his adventure, Mike asks you to find the minimum amount of energy required to reach each of the intersections from his home. Formally, for each 1 ≤ i ≤ n Mike is interested in finding minimum possible total energy of some sequence p1 = 1, p2, ..., pk = i.

Input

The first line contains an integer n (1 ≤ n ≤ 200 000) — the number of Mike's city intersection.

The second line contains n integers a1, a2, ..., an (i ≤ ai ≤ n , , describing shortcuts of Mike's city, allowing to walk from intersection i to intersection ai using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from ai to i).

Output

In the only line print n integers m1, m2, ..., mn, where mi denotes the least amount of total energy required to walk from intersection 1 to intersection i.

Examples

Input

3
2 2 3

Output

0 1 2 

Input

5
1 2 3 4 5

Output

0 1 2 3 4 

Input

7
4 4 4 4 7 7 7

Output

0 1 2 1 2 3 3 

Note

In the first sample case desired sequences are:

1: 1; m1 = 0;

2: 1, 2; m2 = 1;

3: 1, 3; m3 = |3 - 1| = 2.

In the second sample case the sequence for any intersection 1 < i is always 1, i and mi = |1 - i|.

In the third sample case — consider the following intersection sequences:

1: 1; m1 = 0;

2: 1, 2; m2 = |2 - 1| = 1;

3: 1, 4, 3; m3 = 1 + |4 - 3| = 2;

4: 1, 4; m4 = 1;

5: 1, 4, 5; m5 = 1 + |4 - 5| = 2;

6: 1, 4, 6; m6 = 1 + |4 - 6| = 3;

7: 1, 4, 5, 7; m7 = 1 + |4 - 5| + 1 = 3.

题目大意：

   有n个点，主人公在1号点，现在假设相邻点之间的距离为1（表述为），并且每个点都有一条到其他点的距离为1的捷径（可以是自己，也可以是相邻点，也可以是远处的点，但是这条边是单向边）。定义了一下两点之间最短路的定义，求1号点到其他点的最短距离是多少？

解题报告：

    话说啊作为cf div2的B题，考最短路过分了啊虽然是裸题，，但是作为B题，，代码超50行了过分了啊。。

AC代码：

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
using namespace std;
const int MAX = 2e5 + 5; 
ll dis[MAX];
bool vis[MAX];
int n,cnt;
vector<ll> vv[MAX];
struct Point {int pos;ll c;Point(){}Point(int pos,ll c):pos(pos),c(c){}bool operator < (const Point & b) const{return c > b.c;}
} ;
void Dijkstra() {for(int i = 1; i<=n; i++) dis[i] = 0x3f3f3f3f3f3f;memset(vis,0,sizeof vis);dis[1] = 0;priority_queue<Point> pq;pq.push(Point(1,0));while(!pq.empty()) {Point cur = pq.top();pq.pop();if(vis[cur.pos]) continue;vis[cur.pos] = 1;int up = vv[cur.pos].size();for(int i = 0; i<up; i++) {int v = vv[cur.pos][i];if(vis[v]) continue;if(dis[v] > dis[cur.pos] + 1) {dis[v] = dis[cur.pos] + 1;pq.push(Point(v,dis[v]));}}}
}
int main()
{cin>>n;int tmp;memset(head,-1,sizeof head);for(int i = 1; i<=n; i++) {scanf("%d",&tmp);vv[i].pb(tmp);}for(int i = 1; i<=n; i++) {if(i != 1) vv[i].pb(i-1);if(i != n) vv[i].pb(i+1);}Dijkstra();for(int i = 1; i<=n; i++) {printf("%lld%c",dis[i],i == n ? '\n' : ' ');}return 0 ;}