B. Colourblindness

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has a grid with 22 rows and nn columns. He colours each cell red, green, or blue.

Vasya is colourblind and can't distinguish green from blue. Determine if Vasya will consider the two rows of the grid to be coloured the same.

Input

The input consists of multiple test cases. The first line contains an integer tt (1≤t≤1001≤t≤100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤1001≤n≤100) — the number of columns of the grid.

The following two lines each contain a string consisting of nn characters, each of which is either R, G, or B, representing a red, green, or blue cell, respectively — the description of the grid.

Output

For each test case, output "YES" if Vasya considers the grid's two rows to be identical, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Example

input

Copy

6

2

RG

RB

4

GRBG

GBGB

5

GGGGG

BBBBB

7

BBBBBBB

RRRRRRR

8

RGBRRGBR

RGGRRBGR

1

G

G

output

Copy

YES
NO
YES
NO
YES
YES

Note

In the first test case, Vasya sees the second cell of each row as the same because the second cell of the first row is green and the second cell of the second row is blue, so he can't distinguish these two cells. The rest of the rows are equal in colour. Therefore, Vasya will say that the two rows are coloured the same, even though they aren't.

In the second test case, Vasya can see that the two rows are different.

In the third test case, every cell is green or blue, so Vasya will think they are the same.

解题说明：此题是一道模拟题，只需要判断两行中红色出现的位置是否相同，除了这些位置外不再出现红色即可。

#include
#include
 
int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int i, n, flag = 0;
		scanf("%d", &n);
		char s1[101], s2[101];
		scanf("%s%s", s1, s2);
		for (i = 0; i
		{
			if ((s1[i] == 'R' && s2[i] != 'R') || (s2[i] == 'R' && s1[i] != 'R'))
			{
				flag = 1;
			}
		}
		if (flag == 1)
		{
			printf("NO\n");
		}
		else
		{
			printf("YES\n");
		}
	}
	return 0;
}