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>
const int INF = 1e9;
const int MOD = 1e9+7;
const long long LINF = 1e18;
#define dump(x)  cout << 'x' << ' = ' << (x) << ` `;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
typedef long long ll;
using namespace std;
typedef pair<ll, ll> P;

template<typename T>
void print(const vector<T> &x) {
    int n = x.size();
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x) {
    int n = x.size();
    rep(i,n) {
        rep(j,x[i].size()) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

template<typename T>
void print(const vector<T> &x, int n) {
    rep(i,n) {
        cout << x[i];
        if (i!=n-1) cout<<" ";
        else cout << endl;
    }
}

template<typename T>
void print(const vector<vector<T>> &x, int n, int m) {
    rep(i,n) {
        rep(j,m) {
            cout << x[i][j] << " ";
        }
        cout << endl;
    }
}

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

void input_init() {
    cin.tie(0); ios::sync_with_stdio(false);
}

template <class T>
class SegTree {
    int n; // number of leaves
    vector<T> data; // vector which has datas
    T def; // init number & 単位元
    function<T(T,T)> operation; // 区間クエリで使う処理
    function<T(T,T)> update; // 点更新で使う処理

    T _query(int a, int b, int k, int l, int r) {
        // [a, b)のクエリ
        // ノードk(区間[l, r)担当)が答える
        if (b <= l || r <= a) return def; // 区間がかぶらなければdef
        if (a <= l && r <= b) return data[k]; // 区間が完全にかぶっていればdata[k]
        else {
            // 一部だけかぶる場合は左右の子に委託する
            ll c1 = _query(a, b, 2*k+1, l, (l+r)/2);
            ll c2 = _query(a, b, 2*k+2, (l+r)/2, r);
            return operation(c1, c2);
        }
    }

    public:

        SegTree(size_t _n, T _def, function<T(T,T)> _operation, function<T(T,T)> _update) :
            def(_def), operation(_operation), update(_update){
                n = 1;
                while (n < _n) {n *= 2;}
                data = vector<T>(2*n-1, def);
        }

        T query(T a, T b) {return _query(a,b,0,0,n);}

        void change(T i, T x) {
            i += n-1;
            data[i] = update(data[i],x);
            while (i>0) {
                i = (i-1)/2;
                data[i] = operation(data[i*2+1], data[i*2+2]);
            }
        }

        T operator[](int i) {return data[i+n-1];}
};


int main(int argc, char const *argv[]) {
    int n,q;
    string s; cin>>n>>s;
    SegTree<ll> roq((size_t)n, 0ll, [](ll a, ll b){return a|b;}, [](ll a, ll b){return b;});

    rep(i,n) {
        ll d = s[i]-'a';
        ll bit = 0;
        bit |= (1ll<<d);
        roq.change(i,bit);
    }

    cin>>q;
    while (q--) {
        int com; cin>>com;
        if (com==1) {
            int ind; char c;
            cin>>ind>>c; --ind;
            ll d = c-'a';
            ll bit = 0;
            bit |= (1ll<<d);
            roq.change(ind,bit);
        }
        else {
            ll l,r; cin>>l>>r;
            --l;
            ll ret = roq.query(l,r);
            ll ans = __builtin_popcountll(ret);
            cout << ans << '\n';
        }
    }
    return 0;
}