codeforces A. Transformation: from A to B(水题)

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

YES5
100 200 2001 4002 40021 

小看了一个点~水题wa到想哭。。一直同一个样例傻傻挣扎。。
题意很简单：只有两种操作，一种是对当前数字a * 2，一种是a * 10 + 1，问能不能得到最后的数字b
想法也容易想到，反过来对b进行操作，如果b是偶数，那必定他是 * 2变化而来，本来我理所当然的想了反之，然而各位不是1的奇数就是NO..[抹泪..嘤嘤嘤]...想法就是这样了。。AC 代码如下：

#include<stdio.h>#include<string.h>int c[10000];int main(){int a, b;while (scanf("%d%d", &a, &b) == 2){int i = 0, tmp = b;if (a == b) i = 1;while (a != b&&b){if (b % 2){if (b % 10 == 1) b /= 10;else break;}else b /= 2;c[i++] = b;}printf("%s\n", a == b ? "YES" : "NO");if (a == b) {printf("%d\n%d ", i + 1, a);for (int j = i - 2; j >= 0; j--)printf("%d ", c[j]);printf("%d\n", tmp);}}return 0;}