You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $x$ that occurs in the array $2$ or more times. Take the first two occurrences of $x$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $2\cdot x$).

Determine how the array will look after described operations are performed.For example, consider the given array looks like $[$

3,4,1,2,2,1,1].

It will be changed in the following way: $[3,4,1,2,2,1,1]\to [3,4,2,2,2,1]\to [3,4,4,2,1]\to [3,8,2,1]$.If the given array is look like $[1,1,3,1,1]$,it will be changed in the following way: $[1,1,3,1,1]\to [2,3,1,1]\to [2,3,2]\to [3,4]$.

Input

The first line contains a single integer $n$($2\le n\le 150000$) — the number of elements in the array.

The second line contains a sequence from $n$elements $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$) — the elements of the array.

Output

In the first line print an integer $k$— the number of elements in the array after all the performed operations. In the second line print $k$ integers — the elements of the array after all the performed operations.

ExamplesInput

7
3 4 1 2 2 1 1

Output

4
3 8 2 1 

Input

5
1 1 3 1 1

Output

2
3 4 

Input

5
10 40 20 50 30

Output

5
10 40 20 50 30 

Note

The first two examples were considered in the statement.

In the third example all integers in the given array are distinct, so it will not change.

題意：給你n個數，然後如果同一個數出現過兩次或者多次的話，就把兩個數求和，並放在兩個數的右邊。然後問你最終的那個數列的順序。

解題思路：

一開始看這個題目的時候，我的第一反應就是優先佇列，因為最近剛結束的團隊天梯賽裡就有一道這樣的題，只是稍微變形了求法。

我們把每一個輸入的元素加入下標，如果兩個數相等，那麼就優先輸出最開始輸入的那兩個，如果兩個數不相等，那麼就把最小的那個先記錄下來，然後把另一個壓回到優先佇列中。因為有可能組成另一個稍微大點的值。然後就是兩個數合併過程和輸出記錄過程。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=3e5+10;
typedef struct Node{
	long long sum,pos;
	bool operator < (const Node &a) const {
		if(a.sum==sum)
			return a.pos<pos;
		return a.sum<sum;
	}
}node;
struct query{
	long long sum,pos;
}edge[maxn];
priority_queue<node> qu;
bool cmp(query a,query b){
	return a.pos<b.pos;
}
long long num,n;
int main(){
	int i,j; 
	scanf("%I64d",&n);
	node no,po;
	while(!qu.empty()) qu.pop();
	for(i=1;i<=n;i++){
		scanf("%I64d",&num);
		no.pos=i;no.sum=num;
		qu.push(no);
	}
	long long tot=0;
	while(!qu.empty()){
		if(qu.size()==1){
			no=qu.top();qu.pop();
			edge[tot].sum=no.sum;
			edge[tot++].pos=no.pos;
			break;
		}
		no=qu.top();qu.pop();
		po=qu.top(); qu.pop();
		if(no.sum==po.sum){
			no.sum*=2;
			no.pos=max(no.pos,po.pos);
			qu.push(no);
		}
		else{
			edge[tot].sum=no.sum;
			edge[tot++].pos=no.pos;
			qu.push(po);
		}
	}
	sort(edge,edge+tot,cmp);
	printf("%I64d\n",tot);
	for(i=0;i<tot;i++){
		printf("%I64d",edge[i].sum);
		if(i!=tot-1)
			printf(" ");
	}
	printf("\n");
	return 0;
} 

哎，過了D題也掉分，ORZ。難受。