//#pragma GCC optimize("O3")
// 1200

#include <bits/stdc++.h>
#include <random>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pi;
typedef vector<int> vi; 
typedef vector<pi> vpi;

#define mp make_pair
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rsz resize
#define bk back()
#define pb push_back

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define For(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define Rof(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7;
const ld PI = acos((ld)-1);
mt19937 rng; // or mt19937_64
template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }

void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << h; if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif

#define vl vector<ll>

const int MX = 4e5+5;
ll p=97;
ll pows[MX];

vl hashstr(string S) {
	vl hsh(sz(S)+1);
	For(i,sz(S)) {
		hsh[i+1] = (hsh[i]+(S[i]-'a'+1)*pows[i])%MOD;
	}
	return hsh;
}
void solve() {
	string S;cin>>S;
	int n = sz(S);
	string S2 = S+S;
	pows[0] = 1;
	FOR(i,1,MX) pows[i] = pows[i-1]*p%MOD;

	vector<vl> hsh(2);
	hsh[0] = hashstr(S2);
	reverse(all(S2));
	hsh[1] = hashstr(S2);
	reverse(all(S2));

	vector<map<ll, vpi>> M(2);
	For(k,2) For(i,sz(S)) {
		ll tmp = hsh[k][n+i+1] - hsh[k][i];
		tmp = ((tmp%MOD)+MOD)%MOD;
		tmp = (tmp*pows[n-i])%MOD;
		M[k][tmp].pb({k,i});
	}

	vector<vl> gr(2, vl(n+1));
	int cnt = 1;
	int tmp = -1;
	For(k,2){
		trav(x,M[k]) {
			trav(y,x.s) {
				dbg(cnt,y.f,y.s);
				gr[k][y.s] = cnt;
			}
			cnt++;
		}
		tmp = cnt;
		cnt = 1;
	}
	tmp = max(tmp, cnt);


	int K; cin>>K;
	vpi pre(K+1);
	ll ans = 0;
	vector<vi> grCnt(2,vi(tmp+5));
	grCnt[0][gr[0][0]]++;
	bool hasIrot = true;
	For(i,n/2) if(S[i]!=S[n-1-i]) hasIrot = false;
	For(i,K) {
		char c;cin>>c;
		if(c=='I') {
			pre[i+1] = {1^pre[i].f,-pre[i].s%n+n};
			// pre[i+1] = {1^pre[i].f,pre[i].s%n};
		} else if(c=='R') {
			int x;cin>>x;
			pre[i+1] = {pre[i].f,(pre[i].s+x)%n};
		} else {
			int x;cin>>x;
			pre[i+1] = {pre[i].f,(pre[i].s-x)%n};
		}

		int grId = gr[pre[i+1].f][(pre[i+1].s%n+n)%n];
		dbg(i,grId, ans);
		ans += grCnt[pre[i+1].f][grId];

		if(hasIrot) {
			int grId2 = gr[1^pre[i+1].f][(-pre[i+1].s%n+n)%n];
			ans += grCnt[1^pre[i+1].f][grId2];
		}

		grCnt[pre[i+1].f][grId]++;
	}
	cout << ans << endl;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); 
	solve();
}


/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/