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 N = 2e5 + 5;
bool type[N];
vector<int> values, g[N];
int n, m, lft[N], rgt[N], init_value[N];
long long seg[4 * N], lzy[2][4 * N];
void dfs(int u, int p, bool lvl) {
  type[u] = lvl;
  lft[u] = values.size();
  values.push_back(init_value[u]);
  for (auto &c : g[u])
    if (c != p) dfs(c, u, !lvl);
  rgt[u] = values.size() - 1;
}
void bld(int p, int s, int e) {
  if (s == e) return void(seg[p] = values[s]);
  int m = (s + e) >> 1, a = p << 1, b = a | 1;
  bld(a, s, m);
  bld(b, m + 1, e);
  seg[p] = seg[a] + seg[b];
}
void chk(int p, int s, int e) {
  if (s != e) {
    int a = p << 1, b = a | 1;
    lzy[0][a] += lzy[0][p];
    lzy[0][b] += lzy[0][p];
    lzy[1][a] += lzy[1][p];
    lzy[1][b] += lzy[1][p];
    lzy[0][p] = lzy[1][p] = 0;
  }
}
long long qry(int p, int s, int e, int idx, bool t) {
  chk(p, s, e);
  if (s == e) return lzy[t][p];
  int m = (s + e) >> 1, a = p << 1, b = a | 1;
  return (idx <= m ? qry(a, s, m, idx, t) : qry(b, m + 1, e, idx, t));
}
void upd(int p, int s, int e, int l, int r, int val, bool t) {
  chk(p, s, e);
  if (s > r || e < l) return;
  if (s >= l && e <= r) {
    lzy[t][p] += val;
    lzy[!t][p] -= val;
    chk(p, s, e);
    return;
  }
  int m = (s + e) >> 1, a = p << 1, b = a | 1;
  upd(a, s, m, l, r, val, t);
  upd(b, m + 1, e, l, r, val, t);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> init_value[i];
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1, 0);
  while (m--) {
    int t, x, val;
    cin >> t >> x;
    --x;
    if (t == 1) {
      cin >> val;
      upd(1, 0, n - 1, lft[x], rgt[x], val, type[x]);
    } else
      cout << init_value[x] + qry(1, 0, n - 1, lft[x], type[x]) << '\n';
  }
  return 0;
}
