24 Game

Little X used to play a card game called "24 Game", but recently he has found it too easy. So he invented a new game.

Initially you have a sequence of n integers: 1, 2, ..., n. In a single step, you can pick two of them, let's denote them a and b, erase them from the sequence, and append to the sequence either a + b, or a - b, or a × b.

After n - 1 steps there is only one number left. Can you make this number equal to 24?

Input

The first line contains a single integer n (1 ≤ n ≤ 105).

Output

If it's possible, print "YES" in the first line. Otherwise, print "NO" (without the quotes).

If there is a way to obtain 24 as the result number, in the following n - 1 lines print the required operations an operation per line. Each operation should be in form: "a op b = c". Where a and b are the numbers you've picked at this operation; op is either "+", or "-", or "*"; c is the result of corresponding operation. Note, that the absolute value of c mustn't be greater than 1018. The result of the last operation must be equal to 24. Separate operator sign and equality sign from numbers with spaces.

If there are multiple valid answers, you may print any of them.

Examples

input

Copy

1

output

NO

input

Copy

8

output

YES
8 * 7 = 56
6 * 5 = 30
3 - 4 = -1
1 - 2 = -1
30 - -1 = 31
56 - 31 = 25
25 + -1 = 24

题目大意：输入一个n ，利用到所有的1-n的数，经过+，-，*  的方式最终产生数字 24 。

看了别人的题解，这题只要最终考虑n=5或者n=4的状况（看这代码）  其他数 能够经过 n-(n-1)=1来消掉， 而产生的1  能够经过乘来消掉，每次n-=2，最终n==4或者n==5（由于不知道n是偶数仍是奇数）。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int n;
10     scanf("%d",&n);
11     if(n<=3){
12         printf("NO"); return 0;
13     }
14 
15     printf("YES\n");
16     int cnt=0;
17     while(n>=6)
18     {
19         printf("%d - %d = %d\n",n,n-1,1);
20         cnt++;  //统计 1的个数
21         n-=2;
22         
23     }    //这里结束以后n==4或者n==5
24     if(n==4)
25     {
26         printf("4 * 3 = 12\n");
27         printf("2 * 1 = 2\n");
28         printf("12 * 2 = 24\n");
29     }
30     else //n==5的状况
31     {
32         printf("5 * 4 = 20\n");
33         printf("3 + 2 = 5\n");
34         printf("20 + 5 = 25\n");
35         printf("25 - 1 = 24\n");
36     }
37     while(cnt--)
38         printf("24 * 1 = 24\n"); //最后这个 把上面产生的1 所有用到
39 }