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
n,q=map(int,input().split())
Q=[list(map(int,input().split())) for i in range(q)]


for i in range(30):#要素数以上の2のベキを見つける
    if n<1<<i:#前の問題と違って1-indexなので.
        seg_el=1<<i#Segment treeの台の要素数
        break

SEG=[0 for i in range(2*seg_el-1)]#Segment tree（遅延を用いるが、伝搬が不要。一本のセグ木でOK）

def update(a,b,x,k,l,r):  
    if r<=a or b<=l:#区間[a,b)が対象区間の外にあれば終了
        return 
    if a<=l and r<=b:#区間[a,b)が対象区間の中にあればSEG[k]を更新.後にアクセスされたときに遅延評価する.
        SEG[k]+=x
        return 
    update(a,b,x,k*2+1,l,(l+r)//2)#それ以外のときは,SEG[k*2+1]とSEG[k*2+2]で場合分け
    update(a,b,x,k*2+2,(l+r)//2,r)

def getvalue(n):#値を得る
    i=n+seg_el-1
    ANS=SEG[i]
    i=(i-1)//2
    height=1

    while i>=0:
        ANS+=SEG[i]#親をたどって得た和が答え
        i=(i-1)//2
        height+=1
    return ANS

for query in Q:
    if query[0]==0:
        update(query[1],query[2]+1,query[3],0,0,seg_el)
    else:
        print(getvalue(query[1]))
