Task On The Board

Polycarp wrote on the board a string $s$ containing only lowercase Latin letters (‘a’-‘z’). This string is known for you and given in the input.

After that, he erased some letters from the string $s$, and he rewrote the remaining letters in any order. As a result, he got some new string $t$. You have to find it with some additional information.

Suppose that the string $t$ has length $m$ and the characters are numbered from left to right from $1$ to $m$. You are given a sequence of m integers: $b_1$,$b_2$,$…$,$b_m$, where bi is the sum of the distances $|i−j|$ from the index $i$ to all such indices $j$ that $t_j>t_i$ (consider that ‘a’<‘b’<…<‘z’). In other words, to calculate $b_i$, Polycarp finds all such indices j that the index j contains a letter that is later in the alphabet than $t_i$ and sums all the values $|i−j|$.

For example, if $t =$ “abzb”, then:

since $t_1=$‘a’, all other indices contain letters which are later in the alphabet, that is: $b_1=|1−2|+|1−3|+|1−4|=1+2+3=6$;
since $t_2=$‘b’, only the index $j=$3 contains the letter, which is later in the alphabet, that is: $b_2=|2−3|=1$;
since $t_3=$‘z’, then there are no indexes $j$ such that $t_j>t_i$, thus $b_3=0$;
since $t_4=$‘b’, only the index $j=3$ contains the letter, which is later in the alphabet, that is: $b_4=|4−3|=1$.
Thus, if $t =$“abzb”, then $b=[6,1,0,1]$.

Given the string $s$ and the array $b$, find any possible string $t$ for which the following two requirements are fulfilled simultaneously:

$t$ is obtained from $s$ by erasing some letters (possibly zero) and then writing the rest in any order;
the array, constructed from the string $t$ according to the rules above, equals to the array $b$ specified in the input data.
Input
The first line contains an integer $q (1≤q≤100)$ — the number of test cases in the test. Then $q$ test cases follow.

Each test case consists of three lines:

the first line contains string $s$, which has a length from $1$ to $50$ and consists of lowercase English letters;
the second line contains positive integer $m (1≤m≤|s|)$, where $|s|$ is the length of the string $s$, and $m$ is the length of the array $b$;
the third line contains the integers $b_1,b_2,…,b_m (0≤b_i≤1225)$.
It is guaranteed that in each test case an answer exists.

Output
Output $q$ lines: the $k$-th of them should contain the answer (string $t$) to the $k$-th test case. It is guaranteed that an answer to each test case exists. If there are several answers, output any.

Example

input
4
abac
3
2 1 0
abc
1
0
abba
3
1 0 1
ecoosdcefr
10
38 13 24 14 11 5 3 24 17 0
output
aac
b
aba
codeforces

Note
In the first test case, such strings $t$ are suitable: "aac’, “aab”.

In the second test case, such trings $t$ are suitable: “a”, “b”, “c”.

In the third test case, only the string $t$ equals to “aba” is suitable, but the character ‘b’ can be from the second or third position.
拓撲排序。。

#include<bits/stdc++.h>

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector<int>
#define pii pair<int,int>
#define mii unordered_map<int,int>
#define msi unordered_map<string,int>
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<<x<<endl
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 55;
char a[N];
int b[N], T, n, m, cnt[30], ans[N];
vi tmp1, tmp2;
vi s;

int main() {
    T = qr();
    while (T--) {
        ss(a + 1), n = strlen(a + 1);
        ms(cnt);
        repi(i, 1, n)cnt[a[i] - 'a' + 1]++;
        s.clear();
        repi(i, 1, 26)if (cnt[i])s.pb(i);
        m = qr();
        repi(i, 1, m)b[i] = qr();
        tmp1.clear();
        repi(i, 1, m)if (!b[i])tmp1.pb(i);
        int now = s.size() - 1;
        while (!tmp1.empty()) {
            while (cnt[s[now]] < tmp1.size())now--;
            for (auto it:tmp1)ans[it] = s[now];
            now--;
            tmp2.clear();
            for (auto it:tmp1)
                repi(i, 1, m)if (b[i]) {
                        b[i] -= abs(it - i);
                        if (!b[i])tmp2.pb(i);
                    }
            swap(tmp1, tmp2);
        }
        repi(i, 1, m)pc(ans[i] - 1 + 'a');
        puts("");
    }
    return 0;
}