human
I found an interesting problem on https://codeforces.com/problemset/problem/237/E:
You desperately need to build some string t. For that you've got n more strings s_1, s_2, ..., s_{n}. To build string t, you are allowed to perform exactly |t| (|t| is the length of string t) operations on these strings. Each operation looks like that:  choose any non-empty string from strings s_1, s_2, ..., s_{n}.
 choose an arbitrary character from the chosen string and write it on a piece of paper.
 remove the chosen character from the chosen string. 

Note that after you perform the described operation, the total number of characters in strings s_1, s_2, ..., s_{n} decreases by 1. We are assumed to build string t, if the characters, written on the piece of paper, in the order of performed operations form string t.

There are other limitations, though. For each string s_{i} you know number a_{i} — the maximum number of characters you are allowed to delete from string s_{i}. You also know that each operation that results in deleting a character from string s_{i}, costs i rubles. That is, an operation on string s_1 is the cheapest (it costs 1 ruble), and the operation on string s_{n} is the most expensive one (it costs n rubles).

Your task is to count the minimum amount of money (in rubles) you will need to build string t by the given rules. Consider the cost of building string t to be the sum of prices of the operations you use.

-----Input-----

The first line of the input contains string t — the string that you need to build.

The second line contains a single integer n (1 ≤ n ≤ 100) — the number of strings to which you are allowed to apply the described operation. Each of the next n lines contains a string and an integer. The i-th line contains space-separated string s_{i} and integer a_{i} (0 ≤ a_{i} ≤ 100). Number a_{i} represents the maximum number of characters that can be deleted from string s_{i}.

All strings in the input only consist of lowercase English letters. All strings are non-empty. The lengths of all strings do not exceed 100 characters.

-----Output-----

Print a single number — the...

I tried it in Python, but could not do it. Can you solve it?

 ... Can you provide a Python code to print a single number — the?
assistant
#Here is Python code to print a single number — the.

from heapq import heappush, heappop
class MinCostFlow:
    INF = 10**18

    def __init__(self, N):
        self.N = N
        self.G = [[] for i in range(N)]

    def add_edge(self, fr, to, cap, cost):
        G = self.G
        G[fr].append([to, cap, cost, len(G[to])])
        G[to].append([fr, 0, -cost, len(G[fr])-1])

    def flow(self, s, t, f):
        N = self.N; G = self.G
        INF = MinCostFlow.INF

        res = 0
        H = [0]*N
        prv_v = [0]*N
        prv_e = [0]*N

        while f:
            dist = [INF]*N
            dist[s] = 0
            que = [(0, s)]

            while que:
                c, v = heappop(que)
                if dist[v] < c:
                    continue
                for i, (w, cap, cost, _) in enumerate(G[v]):
                    if cap > 0 and dist[w] > dist[v] + cost + H[v] - H[w]:
                        dist[w] = r = dist[v] + cost + H[v] - H[w]
                        prv_v[w] = v; prv_e[w] = i
                        heappush(que, (r, w))
            if dist[t] == INF:
                return -1

            for i in range(N):
                H[i] += dist[i]

            d = f; v = t
            while v != s:
                d = min(d, G[prv_v[v]][prv_e[v]][1])
                v = prv_v[v]
            f -= d
            res += d * H[t]
            v = t
            while v != s:
                e = G[prv_v[v]][prv_e[v]]
                e[1] -= d
                G[v][e[3]][1] += d
                v = prv_v[v]
        return res

T = input()
need = [0] * 30
for t in T:
    need[ord(t) - ord('a')] += 1
have = [[0] * 30 for _ in range(110)]
N = int(input())
a = [0] * 110

for i in range(N):
    v = input().split()
    S = v[0]
    a[i] = int(v[1])
    for s in S:
        have[i][ord(s)-ord('a')] += 1

V = 30 + N + 10
flow = MinCostFlow(V)

start = 30 + N + 5
target = 30 + N + 6
for c in range(26):
    flow.add_edge(start, c, need[c], 0)
    for j in range(N):
        flow.add_edge(c, 30+j, have[j][c], 0)

for i in range(N):
    flow.add_edge(30+i, target,...