Codeforces Round #732 (Div. 2) B. AquaMoon and Stolen String（思维/字符串）

AquaMoon had 𝑛n strings of length 𝑚m each. 𝑛n is an odd number.

When AquaMoon was gone, Cirno tried to pair these 𝑛n strings together. After making 𝑛−12n−12 pairs, she found out that there was exactly one string without the pair!

In her rage, she disrupted each pair of strings. For each pair, she selected some positions (at least 11 and at most 𝑚m) and swapped the letters in the two strings of this pair at the selected positions.

For example, if 𝑚=6m=6 and two strings "abcdef" and "xyzklm" are in one pair and Cirno selected positions 22, 33 and 66 she will swap 'b' with 'y', 'c' with 'z' and 'f' with 'm'. The resulting strings will be "ayzdem" and "xbcklf".

Cirno then stole away the string without pair and shuffled all remaining strings in arbitrary order.

AquaMoon found the remaining 𝑛−1n−1 strings in complete disarray. Also, she remembers the initial 𝑛n strings. She wants to know which string was stolen, but she is not good at programming. Can you help her?

Input

This problem is made as interactive. It means, that your solution will read the input, given by the interactor. But the interactor will give you the full input at the beginning and after that, you should print the answer. So you should solve the problem, like as you solve the usual, non-interactive problem because you won't have any interaction process. The only thing you should not forget is to flush the output buffer, after printing the answer. Otherwise, you can get an "Idleness limit exceeded" verdict. Refer to the interactive problems guide for the detailed information about flushing the output buffer.

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

The first line of each test case contains two integers 𝑛n, 𝑚m (1≤𝑛≤1051≤n≤105, 1≤𝑚≤1051≤m≤105) — the number of strings and the length of each string, respectively.

The next 𝑛n lines each contain a string with length 𝑚m, describing the original 𝑛n strings. All string consists of lowercase Latin letters.

The next 𝑛−1n−1 lines each contain a string with length 𝑚m, describing the strings after Cirno exchanged and reordered them.

It is guaranteed that 𝑛n is odd and that the sum of 𝑛⋅𝑚n⋅m over all test cases does not exceed 105105.

Hack format:

The first line should contain a single integer 𝑡t. After that 𝑡t test cases should follow in the following format:

The first line should contain two integers 𝑛n and 𝑚m.

The following 𝑛n lines should contain 𝑛n strings of length 𝑚m, describing the original strings.

The following 𝑛−12n−12 lines should describe the pairs. They should contain, in the following order: the index of the first string 𝑖i (1≤𝑖≤𝑛1≤i≤n), the index of the second string 𝑗j (1≤𝑗≤𝑛1≤j≤n, 𝑖≠𝑗i≠j), the number of exchanged positions 𝑘k (1≤𝑘≤𝑚1≤k≤m), and the list of 𝑘k positions that are exchanged (𝑘k distinct indices from 11 to 𝑚m in any order).

The final line should contain a permutation of integers from 11 to 𝑛n, describing the way the strings should be reordered. The strings will be placed in the order indices placed in this permutation, the stolen string index will be ignored.

Output

For each test case print a single line with the stolen string.

Example

input

Copy

3
3 5
aaaaa
bbbbb
ccccc
aaaaa
bbbbb
3 4
aaaa
bbbb
cccc
aabb
bbaa
5 6
abcdef
uuuuuu
kekeke
ekekek
xyzklm
xbcklf
eueueu
ayzdem
ukukuk

output

Copy

ccccc
cccc
kekeke

和交互p关系没有。。

类似异或的思想，对于每个位置的所有字符直接统计数量，到时候找出来谁少了即可。

#include <bits/stdc++.h>
using namespace std;
int n, m;
int cnt[100005][26];
int main() {
	int t;
	cin >> t;
	while(t--) {
		vector<string> v1, v2;
		cin >> n >> m;
		memset(cnt, 0, sizeof(cnt));
		for(int i = 1; i <= n; i++) {
			string tmp;
			cin >> tmp;
			for(int i = 0; i < tmp.size(); i++) {
				cnt[i][tmp[i] - 'a']++;
			}
		}
		for(int i = 1; i <= n - 1; i++) {
			string tmp;
			cin >> tmp;
			v2.push_back(tmp);
			for(int i = 0; i < tmp.size(); i++) {
				cnt[i][tmp[i] - 'a']--;
			}
		}
		for(int i = 0; i < m; i++) {
			for(int j = 0; j < 26; j++) {
				if(cnt[i][j]) {
					cout << (char)('a' + j);
					break;
				}
			}
		} 
		cout << endl;
	}
	return 0;
}