814_C. An impassioned circulation of affection
Nadeko's birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!

Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaint at most m of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c — Brother Koyomi's favourite one, and takes the length of the longest among them to be the Koyomity of the garland.

For instance, let's say the garland is represented by "kooomo", and Brother Koyomi's favourite colour is "o". Among all subsegments containing pieces of "o" only, "ooo" is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.

But problem arises as Nadeko is unsure about Brother Koyomi's favourite colour, and has swaying ideas on the amount of work to do. She has q plans on this, each of which can be expressed as a pair of an integer mi and a lowercase letter ci, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.

Input

The first line of input contains a positive integer n (1 ≤ n ≤ 1 500) — the length of the garland.

The second line contains n lowercase English letters s1s2... sn as a string — the initial colours of paper pieces on the garland.

The third line contains a positive integer q (1 ≤ q ≤ 200 000) — the number of plans Nadeko has.

The next q lines describe one plan each: the i-th among them contains an integer mi (1 ≤ mi ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci — Koyomi's possible favourite colour.

Output

Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.

Examples

Input

6
koyomi
3
1 o
4 o
4 m


Output

3
6
5


Input

15
yamatonadeshiko
10
1 a
2 a
3 a
4 a
5 a
1 b
2 b
3 b
4 b
5 b


Output

3
4
5
7
8
1
2
3
4
5


Input

10
aaaaaaaaaa
2
10 b
10 z


Output

10
10

Note

In the first sample, there are three plans: 

  * In the first plan, at most 1 piece can be repainted. Repainting the "y" piece to become "o" results in "kooomi", whose Koyomity of 3 is the best achievable; 
  * In the second plan, at most 4 pieces can be repainted, and "oooooo" results in a Koyomity of 6; 
  * In the third plan, at most 4 pieces can be repainted, and "mmmmmi" and "kmmmmm" both result in a Koyomity of 5. 
```python
""" Python 3 compatibility tools. """
from __future__ import division, print_function
import itertools
import sys
import os
from io import BytesIO, IOBase


if sys.version_info[0] < 3:
  input = raw_input
  range = xrange

  filter = itertools.ifilter
  map = itertools.imap
  zip = itertools.izip


def is_it_local():
  script_dir = str(os.getcwd()).split('/')
  username = "dipta007"
  return username in script_dir


def READ(fileName):
  if is_it_local():
    sys.stdin = open(f'./{fileName}', 'r')

# region fastio
BUFSIZE = 8192

class FastIO(IOBase):
  newlines = 0

  def __init__(self, file):
    self._fd = file.fileno()
    self.buffer = BytesIO()
    self.writable = "x" in file.mode or "r" not in file.mode
    self.write = self.buffer.write if self.writable else None

  def read(self):
    while True:
      b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
      if not b:
          break
      ptr = self.buffer.tell()
      self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
    self.newlines = 0
    return self.buffer.read()

  def readline(self):
    while self.newlines == 0:
      b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
      self.newlines = b.count(b"\n") + (not b)
      ptr = self.buffer.tell()
      self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
    self.newlines -= 1
    return self.buffer.readline()

  def flush(self):
    if self.writable:
      os.write(self._fd, self.buffer.getvalue())
      self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
  def __init__(self, file):
    self.buffer = FastIO(file)
    self.flush = self.buffer.flush
    self.writable = self.buffer.writable
    self.write = lambda s: self.buffer.write(s.encode("ascii"))
    self.read = lambda: self.buffer.read().decode("ascii")
    self.readline = lambda: self.buffer.readline().decode("ascii")

if not is_it_local():
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion


def input1(type=int):
  return type(input())


def input2(type=int):
  [a, b] = list(map(type, input().split()))
  return a, b


def input3(type=int):
  [a, b, c] = list(map(type, input().split()))
  return a, b, c


def input_array(type=int):
  return list(map(type, input().split()))


def input_string():
  s = input()
  return list(s)

if is_it_local():
  def debug(*args):
    st = ""
    for arg in args:
      st += f"{arg} "
    print(st)
else:
  def debug(*args):
    pass

##############################################################

ltr = [[] for _ in range(26)]
memo = {}


def main():
  n = input1()
  st = input_string()

  for l in range(97, 97+26):
    ltr[l-97].append(0)
    ch = chr(l)
    cum = 0
    for i in range(n):
      if st[i] == ch:
        cum += 1
      ltr[l-97].append(cum)
  
  q = input1()
  for i in range(q):
    [m, c] = list(input().split())
    m = int(m)

    if c in memo and m in memo[c]:
      print(memo[c][m])
      continue
      
    res = m
    low = 1
    z = 0
    l = ord(c) - 97
    for high in range(0, n+1):
      tot = high - low + 1
      now = ltr[l][high] - ltr[l][low-1]
      need = tot - now
      debug(high, low, tot, now, need)
      while need > m:
        low += 1
        tot = high - low + 1
        now = ltr[l][high] - ltr[l][low-1]
        need = tot - now

      res = max(res, high - low + 1)

    if c not in memo:
      memo[c] = {}
    
    memo[c][m] = res
    
    print(res)


  
  pass

if __name__ == '__main__':
  # READ('in.txt')
  main()
```