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 <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<(int)(n);i++)

int N,Q;
string S;
vector<set<int>> pls;
const int INF = 1 << 30;

void solve() {
	pls.assign(26, set<int>({ INF }));
	rep(i, N) pls[S[i] - 'a'].insert(i);

	while (Q--) {
		int type;
		cin >> type;
		if (type == 1) {
			int ind; char nc; cin >> ind >> nc; --ind;
			auto pc = S[ind];
			S[ind] = nc;
			pls[pc - 'a'].erase(ind);
			pls[nc - 'a'].insert(ind);
		}
		else {
			int l, r; cin >> l >> r; --l;
			int res = 0;
			rep(c, 26) {
				if (pls[c].empty()) continue;
				int lid = *pls[c].lower_bound(l);
				int rid = *pls[c].lower_bound(r);
				if (rid > lid) ++res;
			}
			cout << res << endl;
		}
	}
}

int main() {

	cin >> N >> S >> Q;
	solve();

}
