/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
*/
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;
//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}
#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << endl; cout << "\n";}
#define pb push_back
#define umap unordered_map
template<typename T1, typename T2> istream& operator>>(istream& is, pair<T1, T2>& p){is >> p.first >> p.second;return is;}
template<typename T1, typename T2> ostream& operator<<(ostream& os, pair<T1, T2>& p){os <<"" << p.first << " " << p.second << ""; return os;}
void usaco(string taskname){
string fin = taskname + ".in";
string fout = taskname + ".out";
const char* FIN = fin.c_str();
const char* FOUT = fout.c_str();
freopen(FIN, "r", stdin);
freopen(FOUT, "w", stdout);
}
template<typename T>
void read(vector<T>& v){
int n=v.size();
for(int i=0; i<n; i++)
cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
vector<T>ans;
for(T t:a)
if(ans.empty() || t!=ans.back())
ans.push_back(t);
return ans;
}
void solve();
int main(){
GLHF;
int t=1;
//cin >> t;
while(t--)
solve();
}
int dp[int(1e6+1)][5][5][5];
int calc(int l,int bal,int mn_bal,int mx_bal,int mod){
if(dp[l][bal+2][mn_bal+2][mx_bal+2]!=-1)return dp[l][bal+2][mn_bal+2][mx_bal+2];
if(mx_bal-mn_bal>2)return dp[l][bal+2][mn_bal+2][mx_bal+2]=0;
if(!l)
return dp[l][bal+2][mn_bal+2][mx_bal+2]=1;
return dp[l][bal+2][mn_bal+2][mx_bal+2]=(calc(l-1,bal-1,min(mn_bal,bal-1),mx_bal,mod)+calc(l-1,bal+1,mn_bal,max(mx_bal,bal+1),mod))%mod;
}
void solve() {
ll n,mod;
cin >> n >> mod;
string s;
cin >> s;
for(int i=0; i<=n; i++)
for(int j=0; j<5; j++)
for(int k=0; k<5; k++)
for(int l=0; l<5; l++)
dp[i][j][k][l]=-1;
ll ans=0;//how many are lexic. smaller than s
// L < P !!!
int bal=0,mx_bal=0,mn_bal=0;
for(int i=0; i<n; i++) {
if(s[i]=='L'){
bal--;
mx_bal=max(mx_bal,bal);
mn_bal=min(mn_bal,bal);
continue;
}
int cbal=bal,cmn=mn_bal,cmx=mx_bal;
cbal--;
cmn=min(cmn,cbal);
cmx=max(cmx,cbal);
ans = (ans + calc(n-i-1,0,cmn-cbal,cmx-cbal,mod))%mod;
bal++;
mx_bal=max(mx_bal,bal);
mn_bal=min(mn_bal,bal);
}
ans=(ans+1);
out(ans)
}
/*
12
100000000
LPLLPLPPLPLL
*/
