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
"""
E
"""
#treeのiに1を加える
def bit_add(I,k):
    while I<=N:
        tree[k][I]+=1
        I+=I&(-I)

def bit_no_add(I,k):
    while I<=N:
        tree[k][I]-=1
        I+=I&(-I)
        
#0からiまでの合計を帰す
def bit_sum(I,k):
    s=0
    while I>0:
        s+=tree[k][I]
        I-=I&(-I)
    return s

N=int(input())
S=["a"]+list(input())
Q=int(input())
que=[list(input().split()) for i in range(Q)]

alpa=list("abcdefghijklmnopqrstuvwxyz")
tree=[[0]*(N+1) for i in range(len(alpa))]

for i in range(1,N+1):
    j=alpa.index(S[i])
    bit_add(i, j)

for iii in range(Q):
    if que[iii][0]=="1":
        i,c=int(que[iii][1]),que[iii][2]
        j=alpa.index(S[i])
        bit_no_add(i, j)
        S[i]=c
        j=alpa.index(c)
        bit_add(i, j)
    else:
        l,r=que[iii][1],que[iii][2]
        l=int(l)
        r=int(r)
        cnt=0
        for j in range(len(alpa)):
            if bit_sum(l-1, j)==bit_sum(r, j):
                continue
            cnt+=1
        print(cnt)