CodeForces--TechnoCup--2016.10.15--ProblemA--Transformation: from A to B

http://codeforces.com/contest/727/problem/A

Transformation: from A to B

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasily has a number a, which he wants to turn into a number b. For this purpose, he can do two types of operations:

• multiply the current number by 2 (that is, replace the number x by 2·x);
• append the digit 1 to the right of current number (that is, replace the number x by 10·x + 1).

You need to help Vasily to transform the number a into the number b using only the operations described above, or find that it is impossible.

Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform a into b.

Input

The first line contains two positive integers a and b (1 ≤ a < b ≤ 10⁹) — the number which Vasily has and the number he wants to have.

Output

If there is no way to get b from a, print "NO" (without quotes).

Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer k — the length of the transformation sequence. On the third line print the sequence of transformations x₁, x₂, ..., x_k, where:

• x₁ should be equal to a,
• x_k should be equal to b,
• x_i should be obtained from x_i - 1 using any of two described operations (1 < i ≤ k).

If there are multiple answers, print any of them.

Examples

Input

2 162

Output

YES
5
2 4 8 81 162 

Input

4 42

Output

NO

Input

100 40021

Output

YES5100 200 2001 4002 40021 

看到这道题的第一眼,我想到的是用DFS搜索解决,当然搜索完全可以解决.但交上去就出问题了--空间不足--无论怎么优化空间复杂度都会爆栈
最后Lzy突然想到了规律:若某个数是奇数,必然是通过*10+1变来的,而如果某个数是偶数,一定是通过*2得来的.
这样,从b往前倒推,若能得到a,则说明可以,否则不行

#include<iostream>
#include<vector>
#include<stdio.h>
#include<string.h>
#include<stack>
using namespace std;
const int MAXN=;
unsigned int aa[MAXN];
long long a,b;
long long n;
bool flag;
int main()
{
    //freopen("data.in","r",stdin);
    while(cin>>a>>b) {
        flag=;
        n=;
        aa[]=b;
        while(b>) {
            if(b==a) {
                flag=;
                break;
            }
            if(b%) {
                n++;
                b=b-;
                if(b%)
                    break;
                b=(b/);
                aa[n]=b;
            } else {
                n++;
                b=b/;
                aa[n]=b;
            }
        }
        if(flag) {
            cout<<"YES"<<endl;
            cout<<n<<endl;
            for(int i=n; i>=; i--) {
                cout<<aa[i]<<" ";
            }
            cout<<endl;
        } else cout<<"NO"<<endl;
    }
}