Problem: You are given a binary array A=(A_1,A_2,\cdots,A_N) of length N.

Process Q queries of the following types. The i-th query is represented by three integers T_i,L_i,R_i.

* T_i=1: Replace the value of A_j with 1-A_j for each L_i \leq j \leq R_i.
* T_i=2: Calculate the inversion(*) of the array A_{L_i},A_{L_i+1},\cdots,A_{R_i}．



Note：The inversion of the array x_1,x_2,\cdots,x_k is the number of the pair of integers i,j with 1 \leq i < j \leq k, x_i > x_j.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 0 \leq A_i \leq 1
* 1 \leq Q \leq 2 \times 10^5
* 1 \leq T_i \leq 2
* 1 \leq L_i \leq R_i \leq N
* All values in Input are integer.

Input

Input is given from Standard Input in the following format:


N Q
A_1 A_2 \cdots A_N
T_1 L_1 R_1
T_2 L_2 R_2
\vdots
T_Q L_Q R_Q


Output

For each query with T_i=2, print the answer.

Example

Input

5 5
0 1 0 0 1
2 1 5
1 3 4
2 2 5
1 1 3
2 1 2


Output

2
0
1

Solution (in PYTHON3):

Solution:
class lazy_segtree:
  #遅延評価セグメント木
  def __init__(s, op, e, mapping, composition, id, v):
    if type(v) is int: v = [e()] * v
    s._n = len(v)
    s.log = s.ceil_pow2(s._n)
    s.size = 1 << s.log
    s.d = [e()] * (2 * s.size)
    s.lz = [id()] * s.size
    s.e = e
    s.op = op
    s.mapping = mapping
    s.composition = composition
    s.id = id
    for i in range(s._n): s.d[s.size + i] = v[i]
    for i in range(s.size - 1, 0, -1): s.update(i)
  
  # 1点更新
  def set(s, p, x):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    s.d[p] = x
    for i in range(1, s.log + 1): s.update(p >> i)
 
  # 1点取得
  def get(s, p):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    return s.d[p]
 
  # 区間演算
  def prod(s, l, r):
    if l == r: return s.e()
    l += s.size
    r += s.size
    for i in range(s.log, 0, -1):
      if (((l >> i) << i) != l): s.push(l >> i)
      if (((r >> i) << i) != r): s.push(r >> i)
    sml, smr = s.e(), s.e()
    while (l < r):
      if l & 1: 
        sml = s.op(sml, s.d[l])
        l += 1
      if r & 1:
        r -= 1
        smr = s.op(s.d[r], smr)
      l >>= 1
      r >>= 1
    return s.op(sml, smr)
 
  # 全体演算
  def all_prod(s): return s.d[1]
 
  # 1点写像
  def apply(s, p, f):
    p += s.size
    for i in range(s.log, 0, -1): s.push(p >> i)
    s.d[p] = s.mapping(f, s.d[p])
    for i in range(1, s.log + 1): s.update(p >> i)
 
  # 区間写像
  def apply(s, l, r, f):
    if l == r: return
    l += s.size
    r += s.size
    for i in range(s.log, 0, -1):
      if (((l >> i) << i) != l): s.push(l >> i)
      if (((r >> i) << i) != r): s.push((r - 1) >> i)
    l2, r2 = l, r
    while l < r:
      if l & 1: 
        sml = s.all_apply(l, f)
        l += 1
      if r & 1:
        r -= 1
        smr = s.all_apply(r, f)
      l >>= 1
      r >>= 1
    l, r = l2, r2
    for i in range(1, s.log + 1):
      if (((l >> i) << i) != l): s.update(l >> i)
      if (((r >> i) << i) != r): s.update((r - 1) >> i)
   
  # L固定時の最長区間のR
  def max_right(s, l, g):
    if l == s._n: return s._n
    l += s.size
    for i in range(s.log, 0, -1): s.push(l >> i)
    sm = s.e()
    while True:
      while (l % 2 == 0): l >>= 1
      if not g(s.op(sm, s.d[l])):
        while l < s.size:
          s.push(l)
          l = 2 * l
          if g(s.op(sm, s.d[l])):
            sm = s.op(sm, s.d[l])
            l += 1
        return l - s.size
      sm = s.op(sm, s.d[l])
      l += 1
      if (l & -l) == l: break
    return s._n
 
  # R固定時の最長区間のL
  def min_left(s, r, g):
    if r == 0: return 0
    r += s.size
    for i in range(s.log, 0, -1): s.push((r - 1) >> i)
    sm = s.e()
    while True:
      r -= 1
      while r > 1 and (r % 2): r >>= 1
      if not g(s.op(s.d[r], sm)):
        while r < s.size:
          s.push(r)
          r = 2 * r + 1
          if g(s.op(s.d[r], sm)):
            sm = s.op(s.d[r], sm)
            r -= 1
        return r + 1 - s.size
      sm = s.op(s.d[r], sm)
      if (r & - r) == r: break
    return 0
 
  def update(s, k): s.d[k] = s.op(s.d[2 * k], s.d[2 * k + 1])
  def all_apply(s, k, f):
    s.d[k] = s.mapping(f, s.d[k])
    if k < s.size: s.lz[k] = s.composition(f, s.lz[k])
  def push(s, k):
    s.all_apply(2 * k, s.lz[k])
    s.all_apply(2 * k + 1, s.lz[k])
    s.lz[k] = s.id()
  def ceil_pow2(s, n):
    x = 0
    while (1 << x) < n: x += 1
    return x

def op(l, r):
  return [l[0] + r[0], l[1] + r[1], l[2] + r[2] + l[1] * r[0]]
 
def e():
  return [0, 0, 0]
 
def mapping(l, r):
  if (not l): return r
  return [r[1], r[0], r[1] * r[0] - r[2]]
 
def composition(l, r):
  return (l and not r) or (not l and r)
 
def id():
  return False
 
def g(x):
  return x[2] <= 1
 
import array
N, Q = list(map(int, input().split()))
a = [(0, 1, 0) if A == "1" else (1, 0, 0) for A in input().split()]

seg = lazy_segtree(op, e, mapping, composition, id, a)
ans = []
for i in range(Q):
  t, l, r = list(map(int, input().split()))
  l -= 1
  if t == 1:
    seg.apply(l, r, True)
  else:
    ans.append(seg.prod(l, r)[2])

for i in ans:
  print(i)