A

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 ton, in the order from the highest to the lowest, the volume of thei-th vessel is a_i liters.

Initially, all the vessels are empty. In some vessels water is poured. All the water that overflows from thei-th vessel goes to the(i + 1)-th one. The liquid that overflows from then-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_i liters of water to thep_i-th vessel;
2. Print the number of liters of water in the k_i-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.

Input

The first line contains integer n — the number of vessels (1 ≤ n ≤ 2·10⁵). The second line containsn integersa₁, a₂, ..., a_n — the vessels' capacities (1 ≤ a_i ≤ 10⁹). The vessels' capacities do not necessarily increase from the top vessels to the bottom ones (see the second sample). The third line contains integerm — the number of queries (1 ≤ m ≤ 2·10⁵). Each of the nextm lines contains the description of one query. The query of the first type is represented as "1 p_i x_i", the query of the second type is represented as "2 k_i" (1 ≤ p_i ≤ n,1 ≤ x_i ≤ 10⁹,1 ≤ k_i ≤ n).

Output

For each query, print on a single line the number of liters of water in the corresponding vessel.

Example

Input

25 1061 1 42 11 2 51 1 42 12 2

Output

458

Input

35 10 861 1 122 21 1 61 3 22 22 3

Output

7105

、

思路： 暴力试了多次都过不了 最后选择并查集 好吧 我不会写诗copy网上的

主要思路就是说 当第输入的小船满了 之后直接加到后面没满的小船上这样子时间会比较少

代码：

    #include<stdio.h>  
    #include<string.h>  
    using namespace std;  
    int n;  
    int sum[1500000];  
    int a[15000000];  
    int f[15000000];  

int find(int a)
{
    int r=a;
    while(r!=f[r])
    r=f[r];
    int i=a;
    int j;
    while(i!=r)
    {
       j=f[i];
       f[i]=r;
       i=j;    
    }
    return r;
}
void merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx>fy)
    f[fy]=fx;
    else
    f[fx]=fy;
}
 
    int main()  
    {  
        while(~scanf("%d",&n))  
        {  
            memset(a,0,sizeof(a));  
            memset(sum,0,sizeof(sum));  
            for(int i=1;i<=n;i++)  
            {  
                scanf("%d",&a[i]);  
                f[i]=i;  
            }  
            f[n+1]=n+1;  
            int q;  
            scanf("%d",&q);  
            while(q--)  
            {  
                int op;  
                scanf("%d",&op);  
                if(op==1)  
                {  
                    int x,have;  
                    scanf("%d%d",&x,&have);
                    while(have>0)
                    {
                     int F=find(x);
                     if(F==n+1)break;
                     if(a[F]-sum[F]>=have)
                     {
                         sum[F]+=have;
                         have=0;
                     }
                     else
                     {
                     have=have-(a[F]-sum[F]);
                     sum[F]=a[F];
                     merge(F,F+1);    
                     }
                    x=F;    
                    }  
                  
                }  
                if(op==2)  
                {  
                    int x;  
                    scanf("%d",&x);  
                    printf("%d\n",sum[x]);  
                }  
            }  
        }  
        return 0;
    }