Problem:
Score : 500 points

Problem Statement

You are given a string S of length N consisting of lowercase English letters.

Process Q queries of the following two types:

Type 1: change the i_q-th character of S to c_q. (Do nothing if the i_q-th character is already c_q.)

Type 2: answer the number of different characters occurring in the substring of S between the l_q-th and r_q-th characters (inclusive).

Constraints

N, Q, i_q, l_q, and r_q are integers.

S is a string consisting of lowercase English letters.

c_q is a lowercase English letter.

1 \leq N \leq 500000

1 \leq Q \leq 20000

|S| = N

1 \leq i_q \leq N

1 \leq l_q \leq r_q \leq N

There is at least one query of type 2 in each testcase.

Input

Input is given from Standard Input in the following format:

N
S
Q
Query_1
\vdots
Query_Q

Here, Query_i in the 4-th through (Q+3)-th lines is one of the following:

1 i_q c_q

2 l_q r_q

Output

For each query of type 2, print a line containing the answer.

Sample Input 1

7
abcdbbd
6
2 3 6
1 5 z
2 1 1
1 4 a
1 7 d
2 1 7

Sample Output 1

3
1
5

In the first query, cdbb contains three kinds of letters: b , c , and d, so we print 3.

In the second query, S is modified to abcdzbd.

In the third query, a contains one kind of letter: a, so we print 1.

In the fourth query, S is modified to abcazbd.

In the fifth query, S does not change and is still abcazbd.

In the sixth query, abcazbd contains five kinds of letters: a, b, c, d, and z, so we print 5.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=m-1; i>=n; i--)
#define SORT(v,n) do{sort(v,v+n);reverse(v,v+n);}while(0)
#define all(x) (x).begin(),(x).end()
#define F first
#define S second
#define MP make_pair
#define MT make_tuple
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
typedef pair<int,int> P;
typedef pair<ll,ll> LP;
 
std::ostream &operator<<(std::ostream &dest, __int128_t value) {
  std::ostream::sentry s(dest);
  if (s) {
    __uint128_t tmp = value < 0 ? -value : value;
    char buffer[128];
    char *d = std::end(buffer);
    do {
      --d;
      *d = "0123456789"[tmp % 10];
      tmp /= 10;
    } while (tmp != 0);
    if (value < 0) {
      --d;
      *d = '-';
    }
    int len = std::end(buffer) - d;
    if (dest.rdbuf()->sputn(d, len) != len) {
      dest.setstate(std::ios_base::badbit);
    }
  }
  return dest;
}

ll POW(ll x,ll n){
  x%=MOD;
  if(n==0)return 1;
  if(n%2==0)return POW(x*x,n/2)%MOD;
  return x%MOD*POW(x,n-1)%MOD;
}
 
ll POW2(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW2(x*x,n/2);
  return x*POW2(x,n-1);
}
 
ll POW3(ll x,ll n,ll m){
  x%=m;
  if(n==0)return 1;
  if(n%2==0)return POW3(x*x,n/2,m)%m;
  return x*POW3(x,n-1,m)%m;
}
 
ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}
 
ll lcm(ll u, ll v) {
  return u/gcd(u,v)*v;
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll KAI2(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI2(m-1);
}
 
ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return m*modinv(n)%MOD*COM(m-1,n-1)%MOD;
}
 
ll COM2(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI2(m)/KAI2(n)/KAI2(m-n);
}
 
ll DEC(ll x,ll m,ll n)//xのm進数でのx^nの位の値
{
  //if(m==2){
    //if(x&(1<<n))return 1;
    //else return 0;
  //}
  return x%POW2(m,n+1)/POW2(m,n);
}
 
ll keta(ll x,ll n)//xのn進数での桁数
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}
 
ll DIV(ll x,ll n)//x!のnで割り切れる回数
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}
 
ll ORD(ll x,ll n)//xのnで割り切れる回数
{
  if(x==0)return INF;
  if(x%n!=0)return 0;
  return 1+ORD(x/n,n);
}
 
ll SUP(ll x,ll n)//xのnで割れなくなるまで割ったときの余り
{
  if(x==0)return 0;
  if(x%n!=0)return x;
  return SUP(x/n,n);
}
 
ll DigSum(ll n)//10進数での桁和
{
  if(n==0)return 0;
  return n%10+DigSum(n/10);
}
 
ll SGS(ll x,ll y, ll m)//1+x+…+x^(y-1)をmで割った余り
{
  if(y==0)return 0;
  if(y%2==0){
    return (1+POW3(x,y/2,m))*SGS(x,y/2,m)%m;
  }
  return (1+x*SGS(x,y-1,m))%m;
}
 
ll SSGS(ll x,ll y,ll m)//Σ[k=1→y](1+x+…+x^(k-1))をmで割った余り
{
  if(y==0)return 0;
  if(y==1)return 1;
  if(y%2==0){
    return (SSGS(x,y/2,m)*(POW3(x,y/2,m)+1)%m+SGS(x,y/2,m)*y/2%m)%m;
  }
  return (SSGS(x,y-1,m)*x%m+y)%m;
}
 
void shuffle(ll array[], ll size) {
    for(ll i = 0; i < size; i++) {
        ll j = rand()%size;
        ll t = array[i];
        array[i] = array[j];
        array[j] = t;
    }
}
 
ll SQRT(ll n){
  ll ok,ng,mid;
  ng=n+1;
  if(303700500<ng)ng=303700500;
  ok=0;
  while(abs(ok-ng)>1){
    mid=(ok+ng)/2;
    if(mid*mid<=n){
      ok=mid;
    }
    else{
      ng=mid;
    }
  }
  return ok;
}
 
struct UnionFind
{
  vector<int> par;
  vector<int> sizes;
  UnionFind(int n) : par(n), sizes(n, 1) {
    rep(i,n) par[i] = i;
  }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (sizes[x] < sizes[y]) swap(x, y);
    par[y] = x;
    sizes[x] += sizes[y];
  }
  bool same(int x, int y) {
    return find(x) == find(y);
  }
  int size(int x) {
    return sizes[find(x)];
  }
};
 
map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
 
bool is_prime(int64_t x) {
  if(x==1)return false;
  for(int64_t i = 2; i * i <= x; i++) {
    if(x % i == 0) return false;
  }
  return true;
}
 
struct edge{ll to, cost;};
struct graph{
  ll V;
  vector<vector<edge> > G;
  vector<ll> d;
 
  graph(ll n){
    init(n);
  }
 
  void init(ll n){
    V = n;
    G.resize(V);
    d.resize(V);
    rep(i,V){
      d[i] = INF;
    }
  }
 
  void add_edge(ll s, ll t, ll cost){
    edge e;
    e.to = t, e.cost = cost;
    G[s].push_back(e);
  }
 
  void dijkstra(ll s){
    rep(i,V){
      d[i] = INF;
    }
    d[s] = 0;
    priority_queue<LP,vector<LP>, greater<LP> > que;
    que.push(LP(0,s));
    while(!que.empty()){
      LP p = que.top(); que.pop();
      ll v = p.second;
      if(d[v]<p.first) continue;
      for(auto e : G[v]){
        if(d[e.to]>d[v]+e.cost){
          d[e.to] = d[v]+e.cost;
          que.push(LP(d[e.to],e.to));
        }
      }
    }
  }
};
 
ll d[410][410];
 
void warshall_floyd(ll n){  
  rep(i,n)rep(j,n)rep(k,n)d[j][k]=min(d[j][k],d[j][i]+d[i][k]);
}
 
struct bit{
  ll m;
  vector<ll> b;
  bit(ll i){
    m=i;
    b.resize(m+1);
  }
  ll num(ll i){
    return b[i];
  }
  ll sum(ll i){
    ll s=0;
    while(i>0){ 
      s+=b[i];
      i-=i&-i;
    }
    return s;
  }
  void add(ll i, ll x){
    while(i<=m){
      b[i]+=x;
      i+=i&-i;
    }
  }
};

int main(){
  ll n,q,a[510000],b[510000],u[30][510000]={},d[510000],x[30][510000]={};
  char c[510000];
  string s;
  cin >> n >> s >> q;
  rep(i,q){
    cin >> a[i] >> b[i];
    if(a[i]==1)cin >> c[i];
    else cin >> d[i];
  }
  rep(i,n)u[s[i]-'a'][i+1]++;
  rep(i,26){
    rep(j,n)u[i][j+1]+=u[i][j];
  }
  rep(i,q){
    if(a[i]==1)continue;
    rep(j,26)x[j][i]=u[j][d[i]]-u[j][b[i]-1];
  }
  rep(i,q){
    ll t=0;
    char g,h;
    if(a[i]==2){
      rep(j,26)if(x[j][i]>0)t++;
      printf("%lld\n",t);
    }
    else{
      g=s[b[i]-1];
      h=c[i];
      s[b[i]-1]=h;
      FOR(j,i+1,q){
        if(a[j]==1)continue;
        if(b[j]<=b[i]&&b[i]<=d[j])x[g-'a'][j]--;
        if(b[j]<=b[i]&&b[i]<=d[j])x[h-'a'][j]++;
      }
    }
  }
}