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>
#define DAU  ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define PLEC return 0;
using namespace std;
set<int> poz[30];
int n, op, x, y, res;
char ch;
string s;
int main() {
    DAU
    cin >> n >> s;
    for (int i = 0; i < n; ++i)
        poz[s[i] - 'a'].emplace(i);
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> op >> x;
        --x;
        if (op == 1) {
            cin >> ch;
            poz[s[x]-'a'].erase(x);
            poz[ch-'a'].emplace(x);
            s[x] = ch;
        }
        else {
            cin >> y;
            res = 0;
            for (int i = 0; i < 26; ++i)
                res += (poz[i].lower_bound(x) != poz[i].lower_bound(y) && !poz[i].empty());
            cout << res << '\n';
        }
    }
    PLEC
}