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<iostream>
#include<vector>
#include<map>
#include<algorithm>
#include<math.h>
#include<string>
#include <iomanip>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<complex>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef int itn;
const ll LINF = 1e18;
const int INF = 1e9;

//マクロ定義
#define vvint(vec,n,m,l) vector<vector<int>> vec(n, vector<int>(m,l));	// lで初期化
#define vvll(vec,n,m,l) vector<vector<ll>> vec(n,vector<ll>(m,l));
#define vint vector<int>
#define pint pair<int,int>
#define rep(i,a) for(int i=0;i<(a);i++)
#define all(x) (x).begin(),(x).end()
#define debug system("pause")				//デバッグ用
#define ret return 0

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

using Graph=vector<vector<ll>>;

#define ketasitei setprecision(15) //15桁表示

const ll MOD = 1000000007;

struct SegmentTree {
private:
	ll n;
	vector<ll> node;
public:
	SegmentTree(vector<ll> v) {
		ll sz = v.size();
		n = 1; while (sz > n) n *= 2;
		node.resize(2 * n - 1, 0);

		for (ll i = 0; i < sz; i++) {
			node[i + n - 1] = v[i];
		}
		for (ll i = n - 2; i >= 0; i--) {
			node[i] = node[i * 2 + 1] | node[i * 2 + 2];
		}
	}

	void update(ll x, ll val) {
		x += n - 1;
		node[x] = val;

		while (x > 0) {
			x = (x - 1) / 2;
			node[x] = node[x * 2 + 1] | node[x * 2 + 2];
		}
	}

	ll getbitsum(ll a, ll b, ll k = 0, ll l = 0, ll r = -1) {
		if (r < 0) r = n;
		if (r <= a || b <= l) return 0;
		if (a <= l && r <= b) return node[k];


		ll vl = getbitsum(a, b, 2 * k + 1, l, (l + r) / 2);
		ll vr = getbitsum(a, b, 2 * k + 2, (l + r) / 2, r);
		return vl | vr;
	}
};


int main(void)

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

	ll n;
	string s;
	cin >> n >> s;

	vector<ll> sl(n);

	rep(i, n)
	{
		sl[i] = 1LL << (ll)(s[i] - 'a');
	}

	ll q;
	cin >> q;

	SegmentTree bisu(sl);

	rep(i, q)
	{
		ll ty;
		cin >> ty;
		if (ty == 1)
		{
			ll point; char ch;
			cin >> point >> ch;

			bisu.update(point - 1, 1 << (ll)(ch - 'a'));
		}
		else
		{
			ll l, r;
			cin >> l >> r;
			cout << __builtin_popcount(bisu.getbitsum(l - 1, r)) << endl;

		}
	}
	return 0;
}