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.

Example

Input

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


Output

3
1
5
#include <iostream>
#include <set>
using namespace std;

int n, q;
char c[500001];
set<int> ct[26];

int main()
{
	cin >> n >> c;
	
	for (int i = 0; i < n; i++) {
		int p = c[i] - 'a';

		ct[p].insert(i);
	}

	cin >> q;

	for (int j = 0; j < q; j++) {
		int w;

		cin >> w;

		if (w == 1) {
			int i;
			string c;

			cin >> i >> c;

			for (int k = 0; k < 26; k++) {
				if (ct[k].find(i - 1) != ct[k].end())
					ct[k].erase(i - 1);

				if (k == c[0] - 'a')
					ct[k].insert(i - 1);
			}
		}
		else if (w == 2) {
			int l, r, m = 0;
			
			cin >> l >> r;

			for (int u = 0; u < 26; u++) {
				if (ct[u].lower_bound(l - 1) != ct[u].end() 
					&& *ct[u].lower_bound(l - 1) < r)
					m++;
			}

			cout << m << endl;
		}
	}
}
