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;
pair<int, int> range[200005];
long long tree[525000];
int num[200005], coeff[200005], fns;
vector<int> edges[200005];
void dfs(int s, int k) {
  if (coeff[s] != 0) return;
  coeff[s] = k;
  range[s].first = fns;
  int i, t, u;
  t = edges[s].size();
  for (i = 0; i < t; i++) {
    u = edges[s][i];
    dfs(u, -k);
  }
  range[s].second = fns;
  fns++;
}
void propagate(int pos) {
  int x = pos * 2;
  tree[x] += tree[pos];
  tree[x + 1] += tree[pos];
  tree[pos] = 0;
}
void update(int pos, int L, int R, int l, int r, int val) {
  if (L > r || l > R) return;
  if (L >= l && R <= r) {
    tree[pos] += val;
    return;
  }
  if (tree[pos] != 0) propagate(pos);
  int x = pos * 2, mid = (L + R) / 2;
  update(x, L, mid, l, r, val);
  update(x + 1, mid + 1, R, l, r, val);
}
long long query(int pos, int L, int R, int k) {
  if (L > k || k > R) return 0;
  if (L == R) return tree[pos];
  if (tree[pos] != 0) propagate(pos);
  int x = pos * 2, mid = (L + R) / 2;
  long long y = query(x, L, mid, k);
  long long z = query(x + 1, mid + 1, R, k);
  return y + z;
}
int main() {
  int n, m, i, j, u, v;
  long long ans;
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  fns = 1;
  dfs(1, 1);
  for (i = 0; i < m; i++) {
    scanf("%d", &j);
    if (j == 1) {
      scanf("%d %d", &u, &v);
      update(1, 1, n, range[u].first, range[u].second, v * coeff[u]);
    } else {
      scanf("%d", &u);
      ans = query(1, 1, n, range[u].second);
      ans = (ans * coeff[u]) + num[u];
      printf("%lld\n", ans);
    }
  }
  return 0;
}
