Write a program which manipulates a sequence A = {a1, a2, . . . , an} with the following operations:

* add(s, t, x): add x to as, as+1, ..., at.
* get(i): output the value of ai.



Note that the initial values of ai (i = 1, 2, . . . , n) are 0.

Constraints

* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* 1 ≤ s ≤ t ≤ n
* 1 ≤ i ≤ n
* 0 ≤ x ≤ 1000

Input


n q
query1
query2
:
queryq


In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, ith query queryi is given in the following format:


0 s t x


or


1 t


The first digit represents the type of the query. '0' denotes add(s, t, x) and '1' denotes get(i).

Output

For each get operation, print the value.

Examples

Input

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


Output

3
5


Input

4 3
1 2
0 1 4 1
1 2


Output

0
1
# Binary Indexed Tree (Fenwick Tree)
class BIT:
  def __init__(self, n):
    self.n = n
    self.data = [0]*(n+1)
    self.el = [0]*(n+1)
  def sum(self, i):
    s = 0
    while i > 0:
      s += self.data[i]
      i -= i & -i
    return s
  def add(self, i, x):
    # assert i > 0
    self.el[i] += x
    while i <= self.n:
      self.data[i] += x
      i += i & -i
  def get(self, i, j=None):
    if j is None:
      return self.el[i]
    return self.sum(j) - self.sum(i-1)

def solve():
  ans = []
  N, Q = map(int, input().split())
  data0 = BIT(N+1)
  data1 = BIT(N+1)
  for i in range(Q):
    p,*y = map(int, input().split())
    if p==0:
      l,r,x = y
      data0.add(l,-x*(l-1))
      data0.add(r+1,x*r)
      data1.add(l,x)
      data1.add(r+1,-x)
    else:
      r = y[0]
      ans.append(data1.sum(r)*r+data0.sum(r)-data1.sum(r-1)*(r-1)-data0.sum(r-1))
  return ans
print(*solve(),sep='\n')


