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>
int take() {
  int n;
  scanf("%d", &n);
  return n;
}
long long takes() {
  long long n;
  scanf("%lld", &n);
  return n;
}
int cas;
using namespace std;
vector<int> v1[200005];
int times, start[200005], finish[200005], a[200005], S[4 * 200005],
    depth[200005];
bool nw[200005];
void dfs(int x) {
  start[x] = ++times;
  nw[x] = 1;
  for (int i = 0; i < v1[x].size(); i++) {
    int v = v1[x][i];
    if (!nw[v]) {
      depth[v] = depth[x] + 1;
      dfs(v);
    }
  }
  finish[x] = times;
}
void update(int x, int n, int v) {
  while (x <= n) {
    S[x] += v;
    x += (x & -x);
  }
}
int query(int pos, int n) {
  int ans = 0;
  while (pos) {
    ans += S[pos];
    pos -= (pos & -pos);
  }
  return ans;
}
int main() {
  int n = take(), m = take(), u, v;
  for (int i = 1; i <= n; i++) a[i] = take();
  for (int i = 1; i <= n - 1; i++) {
    u = take(), v = take();
    v1[u].push_back(v);
    v1[v].push_back(u);
  }
  dfs(1);
  while (m--) {
    int ty = take(), pos = take();
    if (ty == 1) {
      v = take();
      if (depth[pos] % 2) {
        update(start[pos], n, -v);
        update(finish[pos] + 1, n, v);
      } else {
        update(start[pos], n, v);
        update(finish[pos] + 1, n, -v);
      }
    } else {
      int ans = query(start[pos], n);
      if (depth[pos] % 2) ans = -ans;
      printf("%d\n", a[pos] + ans);
    }
  }
  { return 0; };
}
