codeforces-727A

题目连接：http://codeforces.com/contest/727/problem/A

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, ..., xk, where:

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

If there are multiple answers, print any of them.

Examples

input

2 162

output

YES52 4 8 81 162 

input

4 42

output

NO

input

100 40021

output

YES5100 200 2001 4002 40021 

题目大意：给定一个数n，有两种操作，使其变成给定的目标数。

操作1：变为原来的两倍，即变为2*n，操作2：乘10加1，即变为n*10+1。

若能通过这两种操作变为给定的目标数，则输出yes，并输出变化路径。

解题思路：

没有要求最小变化，所以简单dfs即可，路径利用dfs特性储存在数组即可，路径数组大小考虑一下最坏情况即可（一直乘2，数据范围10^9，大约是2^30）

代码如下：

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

];
int x;

bool dfs(ll n,int r, ll s)
{
    if(n==s)
        return true;
    if(n>s)
        return false;
    lu[r]=n;
    r++;
    x=r;
    +,r,s))
        return true;
    ,r,s))
        return true;
    return false;
}

int main()
{
    ll a,b;
    scanf("%lld%lld",&a,&b);
    ,b))
    {
        lu[x]=b;
        x++;
        cout<<"YES"<<endl;
        cout<<x<<endl;
        ;i<x;i++)
            cout<<lu[i]<<" ";
    }
    else
        cout<<"NO"<<endl;
}