#include <iostream>
#include <vector>
using namespace std;
#define int long long
int M, P;
string og;
const int N = 1e5 + 100;
int nxt[N], prv[N];
pair<int, int> mul(pair<int, int> a, pair<int, int> b)
{
return {(1ll * a.first * b.second + 1ll * a.second * b.first) % M, (1ll * a.second * b.second) % M};
}
pair<int, int> add(pair<int, int> a, pair<int, int> b)
{
return {(a.first + b.first) % M, (a.second + b.second) % M};
}
pair<int, int> sub(pair<int, int> a, pair<int, int> b)
{
return {(a.first + M - b.first) % M, (a.second + M - b.second) % M};
}
pair<int, int> solve(int l, int r)
{
// cout << "at " << l << ' ' << r << endl;
vector<pair<int, int>> opr; // operand
vector<char> op; // operation
for (int i = l; i <= r; i++)
{
int p = opr.size();
if (og[i] == '(')
{
opr.push_back(solve(i + 1, nxt[i] - 1));
i = nxt[i];
}
else if ('0' <= og[i] and og[i] <= '9')
{
// digit
int j = i;
int num = 0;
while (j <= r and '0' <= og[j] and og[j] <= '9')
{
num = ((num * 10) + (og[j] - '0')) % M;
j++;
}
opr.push_back({0, num});
i = j - 1;
}
else if (og[i] == 'x')
{
opr.push_back({1, 0});
}
else
{
op.push_back(og[i]);
}
if (op.size() > 0 and op.back() == '*' and og[i] != '*' and opr.size() > p)
{
auto cur = opr.back();
opr.pop_back();
opr.back() = mul(opr.back(), cur);
op.pop_back();
}
}
pair<int, int> qg = {0, 0};
// cout<<"numbers"<<endl;
// for(auto it:opr)cout<<it.first<<' '<<it.second<<endl;
// cout<<"Operations"<<endl;
// for(auto x:op)
// {
// cout<<x<<' ';
// }
// cout<<endl;
qg = opr[0];
for (int i = 1; i < opr.size(); i++)
{
if (op[i - 1] == '+')
{
qg = add(qg, opr[i]);
}
else
{
qg = sub(qg, opr[i]);
}
}
return qg;
}
main()
{
cin >> og;
vector<int> lst;
for (int i = og.size() - 1; i >= 0; i--)
{
if (og[i] == ')')
{
lst.push_back(i);
}
else if (og[i] == '(')
{
nxt[i] = lst.back();
lst.pop_back();
}
}
cin >> P >> M;
pair<int, int> eq = solve(0, og.size() - 1);
// x*eq.first + eq.second == P (mod M)
// x*eq.first
// cout << eq.first << ' ' << eq.second << endl;
P = ((P - eq.second) % M + M) % M;
for (int x = 0; x < M; x++)
{
if (1ll * x * eq.first % M == P)
{
cout << x << endl;
break;
}
}
}
