During the archaeological research in the Middle East you found the traces of three ancient religions: First religion, Second religion and Third religion. You compiled the information on the evolution of each of these beliefs, and you now wonder if the followers of each religion could coexist in peace.

The Word of Universe is a long word containing the lowercase English characters only. At each moment of time, each of the religion beliefs could be described by a word consisting of lowercase English characters.

The three religions can coexist in peace if their descriptions form disjoint subsequences of the Word of Universe. More formally, one can paint some of the characters of the Word of Universe in three colors: 1, 2, 3, so that each character is painted in at most one color, and the description of the i-th religion can be constructed from the Word of Universe by removing all characters that aren't painted in color i.

The religions however evolve. In the beginning, each religion description is empty. Every once in a while, either a character is appended to the end of the description of a single religion, or the last character is dropped from the description. After each change, determine if the religions could coexist in peace.

Input

The first line of the input contains two integers n, q (1 ≤ n ≤ 100 000, 1 ≤ q ≤ 1000) — the length of the Word of Universe and the number of religion evolutions, respectively. The following line contains the Word of Universe — a string of length n consisting of lowercase English characters.

Each of the following line describes a single evolution and is in one of the following formats: 

  * + i c (i ∈ \{1, 2, 3\}, c ∈ \{a, b, ..., z\}: append the character c to the end of i-th religion description. 
  * - i (i ∈ \{1, 2, 3\}) – remove the last character from the i-th religion description. You can assume that the pattern is non-empty. 



You can assume that no religion will have description longer than 250 characters.

Output

Write q lines. The i-th of them should be YES if the religions could coexist in peace after the i-th evolution, or NO otherwise.

You can print each character in any case (either upper or lower).

Examples

Input


6 8
abdabc
+ 1 a
+ 1 d
+ 2 b
+ 2 c
+ 3 a
+ 3 b
+ 1 c
- 2


Output


YES
YES
YES
YES
YES
YES
NO
YES


Input


6 8
abbaab
+ 1 a
+ 2 a
+ 3 a
+ 1 b
+ 2 b
+ 3 b
- 1
+ 2 z


Output


YES
YES
YES
YES
YES
NO
YES
NO

Note

In the first example, after the 6th evolution the religion descriptions are: ad, bc, and ab. The following figure shows how these descriptions form three disjoint subsequences of the Word of Universe:

<image>
1 1
t
+ 2 p
 2 12
aa
+ 1 a
+ 2 a
+ 3 a
- 1
+ 1 a
- 2
+ 2 a
- 3
+ 3 a
+ 2 a
- 1
- 3
 2 10
uh
+ 1 h
+ 2 u
+ 3 h
- 1
- 2
+ 2 h
+ 3 u
- 2
+ 1 u
- 3
 1 1
z
+ 3 z

NO
 YES
YES
NO
YES
NO
YES
NO
YES
NO
NO
NO
YES
 YES
YES
NO
YES
YES
NO
NO
NO
NO
YES
 YES

#ifdef ONLINE_JUDGE
#pragma GCC target("avx2,avx")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using i128 = __int128_t;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
constexpr char ln = '\n';
istream& operator>>(istream& is, __int128_t& x) {
    x = 0;
    string s;
    is >> s;
    int n = int(s.size()), it = 0;
    if (s[0] == '-') it++;
    for (; it < n; it++) x = (x * 10 + s[it] - '0');
    if (s[0] == '-') x = -x;
    return is;
}
ostream& operator<<(ostream& os, __int128_t x) {
    if (x == 0) return os << 0;
    if (x < 0) os << '-', x = -x;
    deque<int> deq;
    while (x) deq.emplace_front(x % 10), x /= 10;
    for (int e : deq) os << e;
    return os;
}
template<class T1, class T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
    return os << "(" << p.first << ", " << p.second << ")";
}
template<class T> 
ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "{";
    for (int i = 0; i < int(v.size()); i++) {
        if (i) os << ", ";
        os << v[i];
    }
    return os << "}";
}
template<class Container> inline int SZ(Container& v) { return int(v.size()); }
template<typename T> inline void UNIQUE(vector<T>& v) { v.erase(unique(v.begin(), v.end()), v.end()); }
template<class T1, class T2> inline bool chmax(T1& a, T2 b) { if (a < b) {a = b; return true ;} return false ;}
template<class T1, class T2> inline bool chmin(T1& a, T2 b) { if (a > b) {a = b; return true ;} return false ;}
inline int topbit(int x) { return x == 0 ? -1 : 31 - __builtin_clz(x); }
inline int topbit(long long x) { return x == 0 ? -1 : 63 - __builtin_clzll(x); }
inline int botbit(int x) { return x == 0 ? 32 : __builtin_ctz(x); }
inline int botbit(long long x) { return x == 0 ? 64 : __builtin_ctzll(x); }
inline int popcount(int x) { return __builtin_popcount(x); }
inline int popcount(long long x) { return __builtin_popcountll(x); }
inline int kthbit(long long x, int k) { return (x>>k) & 1; }
inline constexpr long long TEN(int x) { return x == 0 ? 1 : TEN(x-1) * 10; }
namespace detail {
    template<typename Tp, int Nb>
    auto make_vector(vector<int>& sizes, Tp const& x) {
        if constexpr (Nb == 1) {
            return vector(sizes[0], x);
        } else {
            int size = sizes[Nb-1];
            sizes.pop_back();
            return vector(size, make_vector<Tp, Nb-1>(sizes, x));
        }
    }
}
template<typename Tp, int Nb>
auto make_vector(int const(&sizes)[Nb], Tp const& x = Tp()) {
    vector<int> s(Nb);
    for (int i = 0; i < Nb; i++) s[i] = sizes[Nb-i-1];
    return detail::make_vector<Tp, Nb>(s, x);
}
inline void print() { cout << "\n"; }
template<class T>
inline void print(const vector<T>& v) {
    for (int i = 0; i < int(v.size()); i++) {
        if (i) cout << " ";
        cout << v[i];
    }
    print();
}
template<class T, class... Args>
inline void print(const T& x, const Args& ... args) {
    cout << x << " ";
    print(args...);
}
#ifdef MINATO_LOCAL
inline void debug_out() { cerr << endl; }
template <class T, class... Args>
inline void debug_out(const T& x, const Args& ... args) {
    cerr << " " << x;
    debug_out(args...);
}
#define debug(...) cerr << __LINE__ << " : [" << #__VA_ARGS__ << "] =", debug_out(__VA_ARGS__)
#define dump(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl
#else
#define debug(...) (void(0))
#define dump(x) (void(0))
#endif
struct fast_ios { fast_ios() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

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

    vector<vector<int>> nex(N+1, vector<int>(26,N));
    for (int i = N-1; i >= 0; i--) {
        rep(j,26) nex[i][j] = nex[i+1][j];
        nex[i][S[i]-'a'] = i;
    }

    vector<vector<int>> T(3);

    vector<int> id(3);
    vector<int> used(N);
    auto check=[&]() {
        iota(all(id),0);
        do {
            fill(all(used),0);
            bool ok = true;
            rep(_,3) {
                int u = id[_];
                int cur = 0;
                for (auto c : T[u]) {
                    if (cur>=N) {
                        ok = false;
                        goto END;
                    }
                    int pos = nex[cur][c];
                    while (pos==N or used[pos]) {
                        pos++;
                        if (pos>=N) {
                            ok = false;
                            goto END;
                        }
                        pos = nex[pos][c];
                    }
                    used[pos] = 1;
                    cur = pos+1;
                }
            }
            END:
            if (ok) return true;
        } while (next_permutation(all(id)));
        return false;
    };
    rep(_,Q) {
        char op; cin >> op;
        if (op=='+') {
            int t; cin >> t;
            t--;
            char c; cin >> c;
            T[t].emplace_back(c-'a');
        } else {
            int t; cin >> t;
            t--;
            T[t].pop_back();
        }
        cout << (check() ? "YES" : "NO") << ln;
    }
}
