#include <bits/stdc++.h>
#define ll long long
#define st first
#define nd second
#define pii pair <ll, ll>
#define rep(i, n, m) for (ll i = (n); i <= (m); i ++)
#define rrep(i, n, m) for (ll i = (n); i >= (m); i --)
using namespace std;
namespace P1 {
int val(char ch) {
return ch == ')' ? -1 : 1;
}
bool dp[111][55][55];
ll a[111];
bool check(int len) {
memset(dp, 0, sizeof dp);
dp[0][0][0] = 1;
int mx = len / 2;
rep(i, 0, len - 1) rep(s1, 0, mx) rep(s2, 0, mx)
if (dp[i][s1][s2]) {
if (s1 + a[i + 1] >= 0)
dp[i + 1][s1 + a[i + 1]][s2] = true;
if (s2 + a[i + 1] >= 0)
dp[i + 1][s1][s2 + a[i + 1]] = true;
if (s1 + a[i + 1] >= 0 && s2 + a[i + 1] >= 0)
dp[i + 1][s1 + a[i + 1]][s2 + a[i + 1]] = true;
}
return dp[len][0][0];
}
string base = "RGB";
int dx[3] = {1, 1, 0};
int dy[3] = {0, 1, 1};
void trace(int len, int s1, int s2) {
if (len == 0) return;
rep(i, 0, 3) {
int ns1 = s1 - dx[i] * a[len];
int ns2 = s2 - dy[i] * a[len];
if (dp[len - 1][ns1][ns2]) {
trace(len - 1, ns1, ns2);
cout << base[i];
return;
}
}
}
void query() {
string t;
cin >> t;
rep(i, 0, t.size() - 1)
a[i + 1] = val(t[i]);
if (check(t.size())) {
trace(t.size(), 0, 0);
cout << '\n';
}
else
cout << "impossible\n";
}
void solve() {
int q;
cin >> q;
while (q --) query();
}
};
namespace P2 {
const long long N = 1 << 15;
bool dp[16][15][15];
ll ans[20], a[20];
bool check(int len) {
memset(dp, 0, sizeof dp);
dp[0][0][0] = 1;
int mx = len / 2;
rep(i, 0, len - 1) rep(s1, 0, mx) rep(s2, 0, mx)
if (dp[i][s1][s2]) {
if (s1 + a[i + 1] >= 0)
dp[i + 1][s1 + a[i + 1]][s2] = true;
if (s2 + a[i + 1] >= 0)
dp[i + 1][s1][s2 + a[i + 1]] = true;
if (s1 + a[i + 1] >= 0 && s2 + a[i + 1] >= 0)
dp[i + 1][s1 + a[i + 1]][s2 + a[i + 1]] = true;
}
return dp[len][0][0];
}
void solve() {
rep(len, 1, 15) {
rep(mask, 0, (1 << len) - 1) {
rep(i, 0, len - 1) if (mask >> i & 1)
a[i + 1] = 1;
else
a[i + 1] = -1;
ans[len] += check(len);
}
}
int q, n;
cin >> q;
while (q --) {
cin >> n;
cout << ans[n] << '\n';
}
}
};
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
int P;
cin >> P;
if (P == 1) P1 :: solve(); else P2 :: solve();
return 0;
}
