#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<vpii> vvpii;
typedef vector<vpll> vvpll;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) (int)(x).size()
#define fi first
#define se second
template<class T> bool ckmin(T &a, const T &b) {return a > b ? a = b, 1 : 0;}
template<class T> bool ckmax(T &a, const T &b) {return a < b ? a = b, 1 : 0;}
namespace debug {
void __print(int x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for(auto z : x) cerr << (f++ ? "," : ""), __print(z); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if(sizeof...(v)) cerr << ", "; _print(v...);}
#ifdef ljuba
#define dbg(x...) cerr << "\e[91m" << "LINE(" << __LINE__ << ") -> " << "[" << #x << "] = ["; _print(x)
#else
#define dbg(x...)
#endif
}
using namespace debug;
const char nl = '\n';
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
/*
погледај сампле тестове, пре него што имплементираш (можда имаш погрешну идеју)
уместо да разматраш неке глупе случајеве на папиру, размишљај заправо о задатку
*/
const int mxN = 1e5 + 12;
struct ft {
int a[mxN];
void update(int i, int x) {
for(; i < mxN; i += i&-i) {
a[i] += x;
}
}
int query(int i) {
int r = 0;
for(; i; i -= i&-i) {
r += a[i];
}
return r;
}
}ft;
void print(int n) {
for(int i = 1; i <= n; ++i) {
cout << ft.query(i) << " ";
}
cout << endl;
}
void solve() {
int n, m;
cin >> n >> m;
vi v(n);
for(auto &z : v)
cin >> z;
sort(all(v));
for(int i = 1; i <= n; ++i) {
ft.update(i, v[i-1]);
ft.update(i+1, -v[i-1]);
}
//spaghetti code
while(m--) {
//print(n);
char c;
int x, y;
cin >> c >> x >> y;
if(c == 'F') {
int l = 1, r = n;
int gde = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(ft.query(mid) >= y) {
gde = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
if(gde == -1)
continue;
int prvi = gde;
int drugi = gde + x - 1;
if(drugi >= n) {
ft.update(prvi, 1);
ft.update(n+1, -1);
continue;
}
int maksi = ft.query(drugi);
l = 1, r = n;
gde = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(ft.query(mid) <= maksi) {
gde = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
//gde sada pokazuje na mesto gde je poslednje maksi
l = prvi, r = drugi;
int gde2 = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(ft.query(mid) >= maksi) {
gde2 = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
int ima = drugi - gde2 + 1; //koliko ima u segmentu najvecih
int t2 = gde;
int t1 = gde-ima+1;
ft.update(t1, 1);
ft.update(t2+1, -1);
ft.update(prvi, 1);
ft.update(gde2, -1);
} else {
int l = 1, r = n;
int levo = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(ft.query(mid) >= x) {
levo = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
l = 1, r = n;
int desno = -1;
while(l <= r) {
int mid = (l + r) / 2;
if(ft.query(mid) <= y) {
desno = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
if(min(levo, desno) == -1 || levo > desno) {
cout << 0 << nl;
} else {
cout << desno-levo+1 << nl;
}
}
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int testCases = 1;
//cin >> testCases;
while(testCases--)
solve();
}
