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: 118388061
n=int(input())
nxt=[0]*(n+5)
v=[0]*(n+5)
for i in range(1,n+1):
    nxt[i]=i+1
a=[0]+list(map(int,input().split()))
T=int(input())
for pp in range(T):
    x=list(map(int,input().split()))
    if(x[0]==2):
        print(v[x[1]])
    else:
        u=x[1]
        w=x[2]
        now=u
        while(now<=n and w+v[now]>=a[now]):
            w-=(a[now]-v[now])
            v[now]=a[now]
            now=nxt[now]
        if(now<=n):
            v[now]+=w
        while(u!=now):
            t=u
            u=nxt[u]
            nxt[t]=now;
