B. Queue

There are n walruses standing in a queue in an airport. They are numbered starting from the queue's tail: the 1-st walrus stands at the end of the queue and the n-th walrus stands at the beginning of the queue. The i-th walrus has the age equal to ai.

The i-th walrus becomes displeased if there's a younger walrus standing in front of him, that is, if exists such j (i < j), that a_i> a_j. The displeasure of the i-th walrus is equal to the number of walruses between him and the furthest walrus ahead of him, which is younger than the i-th one. That is, the further that young walrus stands from him, the stronger the displeasure is.

The airport manager asked you to count for each of n walruses in the queue his displeasure.

Input

The first line contains an integer n (2 ≤ n ≤ 10⁵) — the number of walruses in the queue. The second line contains integers a_i(1 ≤ a_i≤ 10⁹).

Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.

Output

Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus's displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.

Examples

   -  - 

    - - - 

  - - - 

 #include <bits/stdc++.h>

 const int INF=0x3f3f3f3f;

 typedef long long LL;

 const double eps =1e-;

 const int mod=1e9+;

 const int maxn=1e5+;

 using namespace std;

 struct node

 {

     int l;

     int r;

     int val;

 }Tr[maxn<<];

 int a[maxn];    //存放数据

 int ans[maxn];    //存放答案 

 void PushUp(int u)

 {

     Tr[u].val=min(Tr[u<<].val,Tr[u<<|].val);

 }

 void Build(int l,int r,int u)

 {

     Tr[u].l=l; Tr[u].r=r; Tr[u].val=;

     if(l==r)

     {

         Tr[u].val=a[l];

         return ;

     }

     int mid=(l+r)>>;

     Build(l,mid,u<<);

     Build(mid+,r,u<<|);

     PushUp(u);

 }

 void Update(int u,int pos)

 {

     int l=Tr[u].l; int r=Tr[u].r;

     if(l==r)

     {

         Tr[u].val=INF;    //pos位置的答案已得出，将其变为INF

         return ;

     }

     int mid=(l+r)>>;

     if(pos<=mid)

         Update(u<<,pos);

     else

         Update(u<<|,pos);

     PushUp(u);    //向上更新最小值

 }

 void Query(int u,int pos)

 {

     int l=Tr[u].l; int r=Tr[u].r;

     if(l==r)

     {

         ans[pos]=l-pos-;    //得出pos位置的答案

         return ;

     }

     int mid=(l+r)>>;

     if(Tr[u<<|].val<a[pos])    //因为是最右边的最小值,先判断右边

         Query(u<<|,pos);

     else

         Query(u<<,pos);

 }

 int main()

 {

     #ifdef DEBUG

     freopen("sample.txt","r",stdin);

     #endif

     int n;

     scanf("%d",&n);

     for(int i=;i<=n;i++)

         scanf("%d",&a[i]);

     Build(,n,);

     for(int i=;i<=n;i++)    //遍历求每一个答案

     {

         if(Tr[].val>=a[i]) ans[i]=-;    //不满足查询条件

         else Query(,i);

         Update(,i);    //更新最小值

     }

     for(int i=;i<=n;i++)

         printf(i==n? "%d\n":"%d ",ans[i]);

     return ;

 }

 #include <iostream>

 #include <cstdio>

 #include <string>

 #include <cstring>

 #include <cmath>

 #include <queue>

 #include <vector>

 #include <set>

 #include <stack>

 #include <map>

 #include <climits>

 using namespace std;

 #define LL long long

 const int INF=0x3f3f3f3f;

 const int MAXN=;

 int a[MAXN],ans[MAXN];

 int x[MAXN],p[MAXN];

 int main()

 {

     int n;

     while(~scanf("%d",&n))

     {

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

         int sum=;

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

         {

             if(sum==||x[sum-]>=a[i])

             {

                 x[sum]=a[i];

                 p[sum++]=i;

                 ans[i]=-;

             }

             else

             {

                 int k,l=,r=sum-;

                 while(l<=r)

                 {

                     int mid=(l+r)>>;

                     if(x[mid]<a[i]) {k=mid;r=mid-;}

                     else l=mid+;

                 }

                 ans[i]=p[k]-i-;

             }

         }

         printf("%d",ans[]);

         for(int i=; i<=n; i++)

             printf(" %d",ans[i]);

         printf("\n");

     }

     return ;

 }

 #include <cstdio>

 #include <cstring>

 #include <vector>

 #include <algorithm>

 #define MAX 100002

 using namespace std;

 int a[MAX],ans[MAX];

 vector<int> v,num;

 int main()

 {

     int n;

     //freopen("data.txt","r",stdin);

     while(~scanf("%d",&n)){

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

         v.clear();

         num.clear();

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

             if(v.size()== || v.back()>=a[i]){

                 v.push_back(a[i]); num.push_back(i);

                 ans[i]=-;

             }else{

                 int j = (lower_bound(v.rbegin(),v.rend(),a[i]) - v.rbegin());

                 j = (int)v.size() - j - ;

                 ans[i] = num[j+] - i - ;

             }

         }

         for(int i=;i<n;i++){

             if(i) printf(" ");

             printf("%d",ans[i]);

         }

         printf("\n");

     }

     return ;

 }

 91B