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 ll long long

int main() {
    ios_base::sync_with_stdio(false);;
    cin.tie(0);
    cout.tie(0); 
ll n,a,b,l,r,f,ans,i;
char c;
ll k1;
cin>>k1;
string s;
vector<set<ll>>v(26);
cin>>s;
for(i=0;i<s.size();i++)
{v[s[i]-'a'].insert(i);
}

cin>>n;
while(n--)
{cin>>a;
if(a==1)
{cin>>b>>c;
v[s[b-1]-'a'].erase(b-1);
v[c-'a'].insert(b-1);
s[b-1]=c;
}
else
{ans=0;
cin>>l>>r;
--l;
--r;
for(f=0;f<26;++f)
{auto it=v[f].lower_bound(l);
if(*it<=r && it!=v[f].end())
{ans+=1;
}
}
cout<<ans<<"\n";
}
}



return 0;
}