Codeforces1311 C. Perform the Combo （前缀和）

Description：

You want to perform the combo on your opponent in one popular fighting game. The combo is the string s consisting of n lowercase Latin letters. To perform the combo, you have to press all buttons in the order they appear in $s$. I.e. if $s="abca"$ then you have to press $'a'$, then $'b'$, $'c'$ and $'a'$ again.

You know that you will spend m wrong tries to perform the combo and during the i-th try you will make a mistake right after $p_i-th$ button $(1≤p_i<n)$ (i.e. you will press first pi buttons right and start performing the combo from the beginning). It is guaranteed that during the $m+1-th$ try you press all buttons right and finally perform the combo.

$I.e.$ if $s="abca"$, $m=2$ and $p=[1,3]$ then the sequence of pressed buttons will be $'a'$ (here you’re making a mistake and start performing the combo from the beginning), $'a', 'b', 'c'$, (here you’re making a mistake and start performing the combo from the beginning), $'a'$ (note that at this point you will not perform the combo because of the mistake), $'b', 'c', 'a'$.

Your task is to calculate for each button (letter) the number of times you’ll press it.

You have to answer t independent test cases.

Input

The first line of the input contains one integer $t (1≤t≤10^4)$ — the number of test cases.

Then t test cases follow.

The first line of each test case contains two integers $n$ and $m (2≤n≤2⋅10^5, 1≤m≤2⋅10^5)$ — the length of s and the number of tries correspondingly.

The second line of each test case contains the string s consisting of n lowercase Latin letters.

The third line of each test case contains m integers $p_1,p_2,…,p_m (1≤p_i<n)$ — the number of characters pressed right during the $i-th$ try.

It is guaranteed that the sum of $n$ and the sum of m both does not exceed $2⋅10^5 (∑n≤2⋅10^5, ∑m≤2⋅10^5)$.

It is guaranteed that the answer for each letter does not exceed $2⋅10^9$.

Output

For each test case, print the answer — $26$ integers: the number of times you press the button ‘a’, the number of times you press the button ‘b’, …, the number of times you press the button ‘z’.

Example

input

3
4 2
abca
1 3
10 5
codeforces
2 8 3 2 9
26 10
qwertyuioplkjhgfdsazxcvbnm
20 10 1 2 3 5 10 5 9 4

output

4 2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 0 9 4 5 3 0 0 0 0 0 0 0 0 9 0 0 3 1 0 0 0 0 0 0 0 
2 1 1 2 9 2 2 2 5 2 2 2 1 1 5 4 11 8 2 7 5 1 10 1 5 2 

Note

The first test case is described in the problem statement. Wrong tries are “a”, “abc” and the final try is “abca”. The number of times you press ‘a’ is 4, ‘b’ is 2 and ‘c’ is 2.

In the second test case, there are five wrong tries: “co”, “codeforc”, “cod”, “co”, “codeforce” and the final try is “codeforces”. The number of times you press ‘c’ is 9, ‘d’ is 4, ‘e’ is 5, ‘f’ is 3, ‘o’ is 9, ‘r’ is 3 and ‘s’ is 1.

解题思路：

本来是直接模拟的没考虑到最坏的情况，当 $m$ 为 $2e5$ 然后 $p_i$ 都是 $1$ 就超了。
用一个 $sum$ 数组来记录前 $i$ 个数中有多少个 $j$ 字符。
到时候直接把出现错误的加上就行了。

AC代码：

const int N = 2e5 + 10;
int t;
int n, m;
int res, len, pre, now;
char s[N];
int p[N];
ll sum[N][30]; //前i个数中有多少个j字符
ll ans[30];
int main()
{
    sd(t);
    while (t--)
    {
        mem(ans, 0);
        sdd(n, m);
        ss(s + 1);
        rep(i, 1, m)
        {
            sd(p[i]);
        }
        rep(i, 1, n)
        {
            rep(j, 0, 25)
            {
                sum[i][j] = sum[i - 1][j];
            }
            sum[i][s[i] - 'a']++;
        }
        rep(i, 1, m)
        {
            rep(j, 0, 25)
            {
                ans[j] += sum[p[i]][j];
            }
        }
        rep(i, 0, 25)
            ans[i] += sum[n][i];
        rep(i, 0, 25)
        {
            printf("%lld%c", ans[i], i == 25 ? '\n' : ' ');
        }
    }
    return 0;
}