You have two binary strings $a$ and $b$ of length $n$. You would like to make all the elements of both strings equal to $0$. Unfortunately, you can modify the contents of these strings using only the following operation:

You choose two indices $l$ and $r$ ($1 \le l \le r \le n$);

For every $i$ that respects $l \le i \le r$, change $a_i$ to the opposite. That is, $a_i := 1 - a_i$;

For every $i$ that respects either $1 \le i < l$ or $r < i \le n$, change $b_i$ to the opposite. That is, $b_i := 1 - b_i$.

Your task is to determine if this is possible, and if it is, to find such an appropriate chain of operations. The number of operations should not exceed $n + 5$. It can be proven that if such chain of operations exists, one exists with at most $n + 5$ operations.


-----Input-----

Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 2 \cdot 10^5$) — the length of the strings.

The second line of each test case contains a binary string $a$, consisting only of characters 0 and 1, of length $n$.

The third line of each test case contains a binary string $b$, consisting only of characters 0 and 1, of length $n$.

It is guaranteed that sum of $n$ over all test cases doesn't exceed $2 \cdot 10^5$.


-----Output-----

For each testcase, print first "YES" if it's possible to make all the elements of both strings equal to $0$. Otherwise, print "NO". If the answer is "YES", on the next line print a single integer $k$ ($0 \le k \le n + 5$) — the number of operations. Then $k$ lines follows, each contains two integers $l$ and $r$ ($1 \le l \le r \le n$) — the description of the operation.

If there are several correct answers, print any of them.


-----Examples-----

Input
5
3
010
101
2
11
10
4
1000
0011
2
10
10
3
111
111
Output
YES
1
2 2
NO
NO
YES
2
1 2
2 2
YES
2
1 1
2 3


-----Note-----

In the first test case, we can perform one operation with $l = 2$ and $r = 2$. So $a_2 := 1 - 1 = 0$ and string $a$ became equal to 000. $b_1 := 1 - 1 = 0$, $b_3 := 1 - 1 = 0$ and string $b$ became equal to 000.

In the second and in the third test cases, it can be proven that it's impossible to make all elements of both strings equal to $0$.

In the fourth test case, we can perform an operation with $l = 1$ and $r = 2$, then string $a$ became equal to 01, and string $b$ doesn't change. Then we perform an operation with $l = 2$ and $r = 2$, then $a_2 := 1 - 1 = 0$ and $b_1 = 1 - 1 = 0$. So both of string $a$ and $b$ became equal to 00.

In the fifth test case, we can perform an operation with $l = 1$ and $r = 1$. Then string $a$ became equal to 011 and string $b$ became equal to 100. Then we can perform an operation with $l = 2$ and $r = 3$, so both of string $a$ and $b$ became equal to 000.

Source: codeforces | Difficulty: MEDIUM
Tags: Implementation

Constructive algorithms


Solutions:
import sys
input = sys.stdin.readline

def print(*args, end='\n', sep=' ') -> None:
	sys.stdout.write(sep.join(map(str, args)) + end)

def m():
	k = int(input())
	for _ in range(k):
		n = int(input())
		a = input()
		b = input()
		ans = []
		xor = int(a, 2) ^ int(b, 2)
		if xor == 2 ** n - 1:
			ans.append((1, n))
			aa = ''
			for i in range(n):
				aa += str(1 - int(a[i]))
			a = aa
			xor = 0
		if xor == 0:
			a = list(a)
			a.append(0)
			b = list(b)
			if a[0] == '1':
				ans.append((1, n))
				ans.append((2, n))
				a[0] = '0'
			f = 0
			p = 0
			for i in range(n + 1):
				if a[i] == '1':
					f = 1
				else:
					if f:
						ans.append((1, i))
						ans.append((1, p + 1))
						f = 0
					p = i
			print('YES')
			print(len(ans))
			for el in ans:
				print(el[0], el[1])
		else:
			print('NO')
m()