There is a system of *n* vessels arranged one above the other as shown in the figure below. Assume that the vessels are numbered from 1 to *n*, in the order from the highest to the lowest, the volume of the *i*-th vessel is *a*<sub class="lower-index">*i*</sub> liters.

Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from the *i*-th vessel goes to the (*i* + 1)-th one. The liquid that overflows from the *n*-th vessel spills on the floor.

Your task is to simulate pouring water into the vessels. To do this, you will need to handle two types of queries:
 1.  Add *x*<sub class="lower-index">*i*</sub> liters of water to the *p*<sub class="lower-index">*i*</sub>-th vessel; 1.  Print the number of liters of water in the *k*<sub class="lower-index">*i*</sub>-th vessel. 
When you reply to the second request you can assume that all the water poured up to this point, has already overflown between the vessels.
The first line contains integer *n* — the number of vessels (1 ≤ *n* ≤ 2·10<sup class="upper-index">5</sup>). The second line contains *n* integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> — the vessels' capacities (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integer *m* — the number of queries (1 ≤ *m* ≤ 2·10<sup class="upper-index">5</sup>). Each of the next *m* lines contains the description of one query. The query of the first type is represented as "1 *p*<sub class="lower-index">*i*</sub> *x*<sub class="lower-index">*i*</sub>", the query of the second type is represented as "2 *k*<sub class="lower-index">*i*</sub>" (1 ≤ *p*<sub class="lower-index">*i*</sub> ≤ *n*, 1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>, 1 ≤ *k*<sub class="lower-index">*i*</sub> ≤ *n*).
# LUOGU_RID: 103854152
def add(x,c):
    global head
    p=head
    while p!=-1 and c!=0:
        if x<=p:
            water=min(v[p]-a[p][0],c)
            a[p][0]+=water
            c-=water
            if a[p][0]==v[p]:
                if p==head:
                    head=a[p][1]
                else:
                    a[q][1]=a[p][1]
        p=a[p][1]
head=0
n=int(input())
v=list(map(int,input().split()))
a=[]
for i in range(n-1):
    a.append([0,i+1])
a.append([0,-1])
q=int(input())
while q>0:
    q-=1
    opt=list(map(int,input().split()))
    if opt[0]==3:
        break
    elif opt[0]==1:
        add(opt[1]-1,opt[2])
    else:
        print(str(a[opt[1]-1][0]))
