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;
int const N = 200200;
int n, m, a, b, t, Enter;
pair<int, int> Seg[10 * N], Lazy[10 * N];
int Depth[N], In[N], Out[N], Arr[N];
vector<int> Tree[N];
void DFS(int u = 1, int p = -1) {
  In[u] = ++Enter;
  for (auto i : Tree[u])
    if (i != p) {
      Depth[i] = Depth[u] + 1;
      DFS(i, u);
    }
  Out[u] = Enter;
}
void PushLazy(int L, int R, int Node) {
  int Range = R - L + 1;
  Seg[Node].first += Range * Lazy[Node].first;
  Lazy[Node << 1].first += Lazy[Node].first;
  Lazy[Node << 1 | 1].first += Lazy[Node].first;
  Lazy[Node].first = 0;
  Seg[Node].second += Range * Lazy[Node].second;
  Lazy[Node << 1].second += Lazy[Node].second;
  Lazy[Node << 1 | 1].second += Lazy[Node].second;
  Lazy[Node].second = 0;
}
void Update(int i, int j, int Val, bool Level, int Node = 1, int L = 1,
            int R = n) {
  PushLazy(L, R, Node);
  if (L > j || R < i) return;
  if (L >= i && R <= j) {
    if (Level) {
      Lazy[Node].first += Val;
      Lazy[Node].second += -Val;
      PushLazy(L, R, Node);
    } else {
      Lazy[Node].first += -Val;
      Lazy[Node].second += Val;
      PushLazy(L, R, Node);
    }
    return;
  }
  int Mid = L + R >> 1;
  Update(i, j, Val, Level, Node << 1, L, Mid);
  Update(i, j, Val, Level, Node << 1 | 1, Mid + 1, R);
}
int Query(int i, int j, int Level, int Node = 1, int L = 1, int R = n) {
  PushLazy(L, R, Node);
  if (L > j || R < i) return 0;
  if (L >= i && R <= j)
    if (Level)
      return Seg[Node].first;
    else
      return Seg[Node].second;
  int Mid = L + R >> 1;
  int Q1 = Query(i, j, Level, Node << 1, L, Mid);
  int Q2 = Query(i, j, Level, Node << 1 | 1, Mid + 1, R);
  return Q1 + Q2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &Arr[i]);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    Tree[a].push_back(b);
    Tree[b].push_back(a);
  }
  DFS();
  for (int i = 1; i <= n; i++)
    if (Depth[i] & 1)
      Update(In[i], In[i], Arr[i], 0);
    else
      Update(In[i], In[i], Arr[i], 1);
  while (m--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &a, &b);
      Update(In[a], Out[a], b, (Depth[a] % 2 == 0));
    } else {
      scanf("%d", &a);
      if (Depth[a] & 1)
        cout << Query(In[a], In[a], 0) << '\n';
      else
        cout << Query(In[a], In[a], 1) << '\n';
    }
  }
  return 0;
}
