Problem:
Score : 400 points

Problem Statement

Takahashi has a string S consisting of lowercase English letters.

Starting with this string, he will produce a new one in the procedure given as follows.

The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:

If T_i = 1: reverse the string S.

If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.

If F_i = 1 : Add C_i to the beginning of the string S.

If F_i = 2 : Add C_i to the end of the string S.

Help Takahashi by finding the final string that results from the procedure.

Constraints

1 \leq |S| \leq 10^5

S consists of lowercase English letters.

1 \leq Q \leq 2 \times 10^5

T_i = 1 or 2.

F_i = 1 or 2, if provided.

C_i is a lowercase English letter, if provided.

Input

Input is given from Standard Input in the following format:

S
Q
Query_1
:
Query_Q

In the 3-rd through the (Q+2)-th lines, Query_i is one of the following:

1

which means T_i = 1, and:

2 F_i C_i

which means T_i = 2.

Output

Print the resulting string.

Sample Input 1

a
4
2 1 p
1
2 2 c
1

Sample Output 1

cpa

There will be Q = 4 operations. Initially, S is a.

Operation 1: Add p at the beginning of S. S becomes pa.

Operation 2: Reverse S. S becomes ap.

Operation 3: Add c at the end of S. S becomes apc.

Operation 4: Reverse S. S becomes cpa.

Thus, the resulting string is cpa.

Sample Input 2

a
6
2 2 a
2 1 b
1
2 2 c
1
1

Sample Output 2

aabc

There will be Q = 6 operations. Initially, S is a.

Operation 1: S becomes aa.

Operation 2: S becomes baa.

Operation 3: S becomes aab.

Operation 4: S becomes aabc.

Operation 5: S becomes cbaa.

Operation 6: S becomes aabc.

Thus, the resulting string is aabc.

Sample Input 3

y
1
2 1 x

Sample Output 3

xy

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define BOUND 27182818284
#define MAT 2

typedef long long ll;
typedef long long int lli;
typedef pair<ll, ll> P;

ll MOD = 1000000007;
const ll INF = (1ll << 60);
const int INFint = (1<<30);

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i, a, b) for(int i=int(a);i<int(b);++i)

template<class T>
bool umax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool umin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}

// gcd
template<typename T>
T gcd(T a, T b) {
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

ll findGCD(vector<ll> arr) {
    ll result = arr[0];
    for (auto a : arr) {
        result = gcd(a, result);
    }
    return result;
}

template<typename T>
T getlcm(T m, T n) {
    // 引数に0がある場合は0を返す
    if ((0 == m) || (0 == n))
        return 0;
    return ((m / gcd(m, n)) * n); // lcm = m * n / gcd(m,n)
}

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    fill((T *) array, (T *) (array + N), val);
}


// v.front() = -BOUND;
// v.back() = BOUND;

//struct edge{
//    int cost, to;
//
//    edge(int in_cost, int in_to){
//        cost=in_cost;
//        to=in_to;
//    }
//    bool operator<(const edge &a) const
//    {
//        return cost > a.cost;
//    }
//};
ll euler_phi(ll n) {
    ll ret = n;
    for (ll i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            ret -= ret / i;
            while (n % i == 0) n /= i;
        }
    }
    if (n > 1) ret -= ret / n;
    return ret;
}

class Combination {
    long long powmod(long long a, long long p) {
        long long ans = 1LL;
        long long mul = a;
        while (p > 0) {
            if ((p & 1) == 1) {
                ans = (ans * mul) % MOD;
            }
            mul = (mul * mul) % MOD;
            p >>= 1;
        }
        return ans;
    }

public:
    int N;
    long long mod;
    vector<long long> fact;
    vector<long long> revfact;

    Combination(int n, long long m) : N(n), mod(m), fact(n + 1), revfact(n + 1) {
        fact[0] = 1;
        for (int i = 1; i <= N; i++) {
            fact[i] = fact[i - 1] * i;
            fact[i] %= mod;
        }

        revfact[N] = powmod(fact[N], mod - 2);
        for (int i = N - 1; i >= 0; i--) {
            revfact[i] = revfact[i + 1] * (i + 1) % mod;
        }
    }

    long long getCombination(int a, int b) {
        if (a < 0 || b < 0) return 0;
        if (b > a)return 0;
        return fact[a] * revfact[b] % mod * revfact[a - b] % mod;
    }
};



struct mint {
    const int mod = 1000000007;
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%mod+mod)%mod){}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
};

struct UnionFind {
    int n, cnt;
    vector<int> par, rank, sz;

    UnionFind(int n) : n(n), cnt(n), par(n), rank(n), sz(n, 1) { iota(par.begin(), par.end(), 0); }

    int find(int x) {
        if (x == par[x]) return x;
        return par[x] = find(par[x]);
    }

    bool same(int x, int y) { return find(x) == find(y); }

    int size(int x) { return sz[find(x)]; }

    void unite(int x, int y) {
        x = find(x), y = find(y);
        if (x == y) return;
        if (rank[x] < rank[y]) {
            par[x] = y;
            sz[y] += sz[x];
        } else {
            par[y] = x;
            sz[x] += sz[y];
            if (rank[x] == rank[y]) {
                rank[x]++;
            }
        }
        cnt--;
    }
};

const string Yes = "Yes";
const string YES = "YES";
const string No = "No";
const string NO = "NO";


long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

template< typename Monoid >
struct SegmentTree {
    using F = function< Monoid(Monoid, Monoid) >;

    int sz;
    vector< Monoid > seg;

    const F f;
    const Monoid M1;

    SegmentTree(int n, const F f, const Monoid &M1) : f(f), M1(M1) {
        sz = 1;
        while(sz < n) sz <<= 1;
        seg.assign(2 * sz, M1);
    }

    void set(int k, const Monoid &x) {
        seg[k + sz] = x;
    }

    void build() {
        for(int k = sz - 1; k > 0; k--) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }

    void update(int k, const Monoid &x) {
        k += sz;
        seg[k] = x;
        while(k >>= 1) {
            seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
        }
    }

    Monoid query(int a, int b) {
        Monoid L = M1, R = M1;
        for(a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if(a & 1) L = f(L, seg[a++]);
            if(b & 1) R = f(seg[--b], R);
        }
        return f(L, R);
    }

    Monoid operator[](const int &k) const {
        return seg[k + sz];
    }

    template< typename C >
    int find_subtree(int a, const C &check, Monoid &M, bool type) {
        while(a < sz) {
            Monoid nxt = type ? f(seg[2 * a + type], M) : f(M, seg[2 * a + type]);
            if(check(nxt)) a = 2 * a + type;
            else M = nxt, a = 2 * a + 1 - type;
        }
        return a - sz;
    }


    template< typename C >
    int find_first(int a, const C &check) {
        Monoid L = M1;
        if(a <= 0) {
            if(check(f(L, seg[1]))) return find_subtree(1, check, L, false);
            return -1;
        }
        int b = sz;
        for(a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
            if(a & 1) {
                Monoid nxt = f(L, seg[a]);
                if(check(nxt)) return find_subtree(a, check, L, false);
                L = nxt;
                ++a;
            }
        }
        return -1;
    }

    template< typename C >
    int find_last(int b, const C &check) {
        Monoid R = M1;
        if(b >= sz) {
            if(check(f(seg[1], R))) return find_subtree(1, check, R, true);
            return -1;
        }
        int a = sz;
        for(b += sz; a < b; a >>= 1, b >>= 1) {
            if(b & 1) {
                Monoid nxt = f(seg[--b], R);
                if(check(nxt)) return find_subtree(b, check, R, true);
                R = nxt;
            }
        }
        return -1;
    }
};

int main() {
    string S; cin >> S;
    int Q; cin >> Q;

    //
    bool flag=true;
    string A, B;
    rep(i, Q){
        int T; cin >> T;

        if(T == 1){
            flag = !flag;
        }else{
            int F; cin >> F;
            char c; cin >> c;
            if(F==1){
                if(flag){
                    A = c+A;
                }else{
                    B = B + c;
                }
            }else{
                if(flag){
                    B = B + c;
                }else{
                    A = c+A;
                }
            }
        }
    }
    if(flag){
        cout << A + S + B << endl;
    }else{
        reverse(A.begin(), A.end());
        reverse(B.begin(), B.end());
        reverse(S.begin(), S.end());
        cout << B + S + A << endl;
    }
    return 0;
}