Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.

This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.

This tree supports two types of queries:

  * "1 x val" — val is added to the value of node x; 
  * "2 x" — print the current value of node x. 



In order to help Iahub understand the tree better, you must answer m queries of the preceding type.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.

Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.

Output

For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.

Examples

Input

5 5
1 2 1 1 2
1 2
1 3
2 4
2 5
1 2 3
1 1 2
2 1
2 2
2 4


Output

3
3
0

Note

The values of the nodes are [1, 2, 1, 1, 2] at the beginning.

Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].

Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].

You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 200010;
int N, M, Type, Val, X, Y, A[NMAX], AibOdd[2 * NMAX], AibEven[2 * NMAX],
    Level[NMAX], First[NMAX], Last[NMAX], K;
vector<int> G[NMAX];
bool Used[NMAX];
void DFS(int Node, int Father) {
  Used[Node] = 1;
  Level[Node] = Level[Father] + 1;
  First[Node] = ++K;
  for (int i = 0; i < G[Node].size(); ++i)
    if (!Used[G[Node][i]]) DFS(G[Node][i], Node);
  Last[Node] = ++K;
}
int LSB(int X) { return (X & (X - 1)) ^ X; }
void Update(int Aib[NMAX], int Pos, int Val) {
  for (; Pos <= 2 * N; Pos += LSB(Pos)) Aib[Pos] += Val;
}
int Query(int Aib[NMAX], int Pos) {
  int Ans = 0;
  for (; Pos; Pos -= LSB(Pos)) Ans += Aib[Pos];
  return Ans;
}
int main() {
  scanf("%i %i", &N, &M);
  for (int i = 1; i <= N; ++i) scanf("%i", &A[i]);
  for (int i = 1; i < N; ++i) {
    scanf("%i %i", &X, &Y);
    G[X].push_back(Y);
    G[Y].push_back(X);
  }
  DFS(1, 0);
  for (; M; M--) {
    scanf("%i", &Type);
    if (Type == 1) {
      scanf("%i %i", &X, &Val);
      if (Level[X] % 2 == 0) {
        Update(AibEven, First[X], Val);
        Update(AibEven, Last[X], -Val);
      } else {
        Update(AibOdd, First[X], Val);
        Update(AibOdd, Last[X], -Val);
      }
    } else {
      scanf("%i", &X);
      if (Level[X] % 2 == 0)
        printf("%i\n",
               A[X] + Query(AibEven, First[X]) - Query(AibOdd, First[X]));
      else
        printf("%i\n",
               A[X] + Query(AibOdd, First[X]) - Query(AibEven, First[X]));
    }
  }
}
