Given a string of length n s = s1, s2,…, sn and m queries. Each query qk (1 ≤ k ≤ m) is one of four types, "L ++", "L-", "R ++", "R-", and l [for the kth query qk. k] and r [k] are defined below.

* L ++: l [k] = l [k-1] + 1, r [k] = r [k-1]

* L-: l [k] = l [k-1] -1, r [k] = r [k-1]

* R ++: l [k] = l [k-1], r [k] = r [k-1] +1

* R-: l [k] = l [k-1], r [k] = r [k-1] -1




However, l [0] = r [0] = 1.

At this time, how many kinds of character strings are created for m substrings sl [k], sl [k] + 1,…, sr [k] -1, sr [k] (1 ≤ k ≤ m). Answer if you can.

Constraints

* The string s consists of a lowercase alphabet

* 1 ≤ n ≤ 3 x 105

* 1 ≤ m ≤ 3 × 105

* qk (1 ≤ k ≤ m) is one of "L ++", "L-", "R ++", "R-"

* 1 ≤ l [k] ≤ r [k] ≤ n (1 ≤ k ≤ m)

Input

Input is given in the following format

> n m
> s
> q1
> q2
>…
> qm
>

Output

Output the solution of the problem on one line

Examples

Input

5 4
abcde
R++
R++
L++
L--


Output

3


Input

4 6
abab
R++
L++
R++
L++
R++
L++


Output

4


Input

10 13
aacacbabac
R++
R++
L++
R++
R++
L++
L++
R++
R++
L--
L--
R--
R--


Output

11
#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT(n) (1LL<<(n))
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end());

typedef long long LL;
typedef unsigned long long ULL;
template<typename T> using V = std::vector<T>;
template<typename T> using VV = std::vector<std::vector<T>>;
template<typename T> using VVV = std::vector<std::vector<std::vector<T>>>;
template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
inline double distance(double y1, double x1, double y2, double x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }

const int INF = 1L << 30;
const double EPS = 1e-9;
const std::string YES = "YES", Yes = "Yes", NO = "NO", No = "No";
const std::vector<int> dy = { 0, 1, 0, -1 }, dx = { 1, 0, -1, 0 };    // 4近傍（右, 下, 左, 上）

using namespace std;


// 部分文字列のhash値を計算
class SubstringHash {
    const int N;
    vector<unsigned long long> powb;
    vector<unsigned long long> shash;
    const unsigned long long b = 1000000007;

public:
    SubstringHash(string s) : N(s.size()) {
        this->powb.resize(s.size() + 10);
        this->shash.resize(s.size() + 10);
        powb[0] = 1;
        FOR(i, 0, s.size()) {
            powb[i + 1] = powb[i] * b;
            shash[i + 1] = shash[i] * b + s[i];
        }
    }

    // s[left, right)のhash値
    unsigned long long hash(int left, int right) {
        assert(0 <= left and 0 <= right and left <= right and right <= N);

        return shash[right] - shash[left] * powb[right - left];
    }
};

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int N, M;
    cin >> N >> M;
    string S;
    cin >> S;

    SubstringHash sh(S);
    set<unsigned long long> hash;
    int left = 0, right = 0;    // 0-index
    FOR(i, 0, M){
        string q;
        cin >> q;
        if      (q == "L++") { left++; }
        else if (q == "L--") { left--; }
        else if (q == "R++") { right++; }
        else if (q == "R--") { right--; }
        hash.insert(sh.hash(left, right + 1));
    }

    print(hash.size());

    return 0;
}
