We have N colored balls arranged in a row from left to right; the color of the i-th ball from the left is c_i.

You are given Q queries. The i-th query is as follows: how many different colors do the l_i-th through r_i-th balls from the left have?

Constraints

* 1\leq N,Q \leq 5 \times 10^5
* 1\leq c_i \leq N
* 1\leq l_i \leq r_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
c_1 c_2 \cdots c_N
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print Q lines. The i-th line should contain the response to the i-th query.

Examples

Input

4 3
1 2 1 3
1 3
2 4
3 3


Output

2
3
1


Input

10 10
2 5 6 5 2 1 7 9 7 2
5 5
2 4
6 7
2 2
7 8
7 9
1 8
6 9
8 10
6 8


Output

1
2
2
1
2
2
6
3
3
3
def segfunc(x,y):
  return x+y
ide_ele=0
class SegTree():
  def __init__(self,init_val,segfunc,ide_ele):
    n=len(init_val)
    self.segfunc=segfunc
    self.ide_ele=ide_ele
    self.num=1<<(n-1).bit_length()
    self.tree=[ide_ele]*2*self.num
    for i in range(n):
      self.tree[self.num+i]=init_val[i]
    for i in range(self.num-1,0,-1):
      self.tree[i]=self.segfunc(self.tree[2*i], self.tree[2*i+1])
  def update(self,k,x):
    k+=self.num
    self.tree[k]=x
    while k>1:
      self.tree[k>>1]=self.segfunc(self.tree[k],self.tree[k^1])
      k>>=1
  def query(self,l,r):
    res=self.ide_ele
    l+=self.num
    r+=self.num
    while l<r:
      if l&1:
        res=self.segfunc(res,self.tree[l])
        l+=1
      if r&1:
        res=self.segfunc(res,self.tree[r-1])
      l>>=1
      r>>=1
    return res
n,q=map(int,input().split())
c=list(map(int,input().split()))
st=SegTree([0]*n,segfunc,ide_ele)
lr=[[]for _ in range(n+1)]
for i in range(q):
  l,r=map(int,input().split())
  lr[r]+=[(l,i)]
d={}
for i in range(n):
  if not d.get(c[i]):d[c[i]]=[i]
  else:d[c[i]]+=[i]
for i in d:
  st.update(d[i][-1],1)
ans=[0]*q
r=n
while r>0:
  if lr[r]:
    l,i=lr[r].pop()
    ans[i]=st.query(l-1,r)
  while not lr[r] and r>0:
    if d[c[r-1]]:st.update(d[c[r-1]].pop(),0)
    if d[c[r-1]]:st.update(d[c[r-1]][-1],1)
    r-=1
for i in ans:print(i)