Vasya has an array of n integers a_1, a_2, …, a_n. Vasya thinks that all numbers in his array are strange for some reason. To calculate how strange the i-th number is, Vasya created the following algorithm.

He chooses a subsegment a_l, a_{l+1}, …, a_r, such that 1 ≤ l ≤ i ≤ r ≤ n, sort its elements in increasing order in his head (he can arrange equal elements arbitrary). After that he finds the center of the segment. The center of a segment is the element at position (l + r) / 2, if the length of the segment is odd, and at position (l + r + 1) / 2 otherwise. Now Vasya finds the element that was at position i before the sorting, and calculates the distance between its current position and the center of the subsegment (the distance between the elements with indices j and k is |j - k|).

The strangeness of the number at position i is the maximum distance among all suitable choices of l and r. 

Vasya wants to calculate the strangeness of each number in his array. Help him to do it. 

Input

The first line contains a single integer n (1 ≤ n ≤ 200 000) — the size of the array.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — Vasya's array.

Output

Print a single line with n numbers. The i-th of them must be equal to the strangeness of the i-th element of the array.

Examples

Input


5
5 4 3 2 1


Output


2 1 1 2 2 


Input


7
3 6 5 6 2 1 3


Output


2 3 1 3 2 3 1 

Note

In the first example:

  1. For the first position we choose the segment from 1 to 5. After sorting, it looks like [1, 2, 3, 4, 5], the center is 3. The distance from the center to 5 is 2.
  2. For the second position we choose the segment from 2 to 4.
  3. For the third position we choose the segment from 3 to 5.
  4. For the fourth position we choose the segment from 1 to 4. After sorting, it looks like [2, 3, 4, 5], the center is 4. The distance from the center to 2 is 2. 
  5. For the fifth position we choose the segment from 1 to 5.
    /**   Created by: Humberto Yusta
          Codeforces User: humbertoyusta
          Country: Cuba                    */
#include<bits/stdc++.h>
using namespace std;
/// Pragmas
#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline","03") // Optimization flags
//#pragma GCC option("arch=native","tune=native","no-zero=upper") // Enable AVX
//#pragma GCC target("avx2") // Enable AVX
//#pragma comment(linker, "/STACK:1024000000,1024000000") // Increase stack limit
//#pragma GCC target("sse,sse2,sse,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") // I don't know what it is
/// Macros:
#define int long long
#define f first
#define s second
#define db(x) cerr << #x << ": " << (x) << '\n';
#define pb push_back
#define all(x) x.begin() , x.end()
#define rall(x) x.rbegin() , x.rend()
typedef pair<int,int> ii;
typedef long double ld;
typedef unsigned long long ull;
typedef long long ll;
/// Constraints:
const int maxn = 200020;
const int mod = 1000000007;
const int mod2 = 998244353;
const ld eps = 1e-9;
const int inf = ((1ll<<31ll)-1ll);
const int INF = 1ll * mod * mod;
const ld pi = acos(-1);
/// Prime Numbers:
// 2, 173, 179, 181, 197, 311, 331, 737, 1009, 2011, 2027, 3079, 4001, 10037, 10079, 20011, 20089;
// 100003, 100019, 100043, 200003, 200017, 1000003, 1000033, 1000037, 1000081;
// 2000003, 2000029, 2000039, 1000000007, 1000000021, 2000000099;
/// Red-Black Tree Template ---------------------------------
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//typedef tree < long long ,  null_type ,  less<long long> ,  rb_tree_tag ,  tree_order_statistics_node_update > ordered_set;
/// Quick Functions------------------------------------------
int qpow(int b,int e){
    if( !e ) return 1;
    if( e & 1 ) return qpow(b,e-1) * b % mod;
    int pwur = qpow(b,e>>1);
    return pwur * pwur % mod;
}
int modinv(int x){ return qpow(x,mod-2); }
bool isprime(int x){
    for(int i=2; i*i<=x; i++)
        if( x % i == 0 )
            return false;
    return true;
}
mt19937_64 r(time(0));
int rng (int a, int b){
    return uniform_int_distribution<int>(a,b)(r);
}
/// My Code -------------------------------------------------

vector<int> v[maxn];

struct node{
    int mn, mx, lazy;
    node(){
        mn = INF;
        mx = -INF;
        lazy = 0;
    }
};

node st[maxn*4];

void reset(){
    for(int i=0; i<maxn*4; i++){
        st[i] = node();
        st[i].mn = 0;
        st[i].mx = 0;
    }
}

node merge_(node a,node b){
    node c;
    c.mn = min( a.mn , b.mn );
    c.mx = max( a.mx , b.mx );
    c.lazy = 0;
    return c;
}

void lazyp(int nod,int l,int r){
    st[nod].mn += st[nod].lazy;
    st[nod].mx += st[nod].lazy;
    if( l != r ){
        st[2*nod].lazy += st[nod].lazy;
        st[2*nod+1].lazy += st[nod].lazy;
    }
    st[nod].lazy = 0;
}

void update(int nod,int l,int r,int x,int y,int val){
    lazyp(nod,l,r);
    if( l > y || r < x ) return;
    if( l >= x && r <= y ){
        st[nod].lazy += val;
        lazyp(nod,l,r);
        return;
    }
    int mi = ( l + r ) >> 1;
        update(2*nod,l,mi,x,y,val);
        update(2*nod+1,mi+1,r,x,y,val);
    st[nod] = merge_( st[2*nod] , st[2*nod+1] );
}

node query(int nod,int l,int r,int x,int y){
    lazyp(nod,l,r);
    if( l > y || r < x ) return node();
    if( l >= x && r <= y ) return st[nod];
    int mi = ( l + r ) >> 1;
    return merge_( query(2*nod,l,mi,x,y) , query(2*nod+1,mi+1,r,x,y) );
}

int32_t main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cout.setf(ios::fixed); cout.precision(10);
    srand(time(NULL));

//    freopen("a.in","r",stdin);
    //freopen("a.in","w",stdout);

    int n;
    cin >> n;
    vector<int> a(n+1);
    vector<int> ans1(n+1);
    vector<int> ans2(n+1);
    for(int i=1; i<=n; i++){
        cin >> a[i];
        v[a[i]].pb(i);
    }

    reset();

    for(int i=1; i<=n; i++)
        update(1,0,n,i,n,-1);
    for(int i=n; i>=1; i--){
        for( auto j : v[i] )
            update(1,0,n,j,n,2);
        for( auto j : v[i] )
            ans1[j] = ( query(1,0,n,j,n).mx - query(1,0,n,0,j).mn ) / 2;
    }

    reset();

    for(int i=1; i<=n; i++)
        update(1,0,n,i,n,-1);
    for(int i=1; i<=n; i++){
        for( auto j : v[i] )
            update(1,0,n,j,n,2);
        for( auto j : v[i] )
            ans2[j] = ( query(1,0,n,j,n).mx - query(1,0,n,0,j).mn - 1 ) / 2;
    }

    for(int i=1; i<=n; i++){
//        db(ans1[i])
//        db(ans2[i])
//        db(max(ans1[i],ans2[i]))
        cout << max( ans1[i] , ans2[i] ) << " \n"[i==n];
    }

    return 0;
}
