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;
template <typename T, typename SS>
ostream& operator<<(ostream& ofs, const pair<T, SS>& p) {
  ofs << "( " << p.first << " , " << p.second << " )";
  return ofs;
}
template <typename T>
void update_max(T last, T& mx) {
  mx = max(mx, last);
}
template <typename T>
void update_min(T last, T& mx) {
  mx = min(mx, last);
}
template <typename T>
void update_sum(T last, T& mx) {
  mx += last;
}
template <typename T>
void print_one(T a) {
  cout << a << ' ';
}
template <typename T, typename... Args>
T max(Args&&... args) {
  T mx = -1e17;
  T dummy[] = {0, ((void)update_max(std::forward<Args>(args), mx), 0)...};
  return mx;
}
template <typename T, typename... Args>
T min(Args&&... args) {
  T mx = 1e17;
  T dummy[] = {0, ((void)update_min(std::forward<Args>(args), mx), 0)...};
  return mx;
}
template <typename T, typename... Args>
T sum(Args&&... args) {
  T mx = 0;
  T dummy[] = {0, ((void)update_sum(std::forward<Args>(args), mx), 0)...};
  return mx;
}
template <typename... Args>
void print(Args&&... args) {
  long long dummy[] = {0, ((void)print_one(std::forward<Args>(args)), 0)...};
  cout << '\n';
}
template <typename T>
void print(vector<T> a) {
  for (auto i : a) cout << i << ' ';
  cout << '\n';
}
template <typename T>
T max(vector<T> a) {
  T mx = -1e17;
  for (auto i : a) mx = max(i, mx);
  return (T)mx;
}
template <typename T>
T min(vector<T> a) {
  T mx = 1e17;
  for (auto i : a) mx = min(i, mx);
  return (T)mx;
}
template <typename T>
T sum(vector<T> a) {
  T mx = 0;
  for (auto i : a) mx += i;
  return mx;
}
vector<vector<long long> > g;
long long n, m;
vector<long long> frst, last;
vector<long long> b;
vector<long long> level;
void dfs(long long v, long long p = -1) {
  frst[v] = b.size();
  b.push_back(v);
  for (auto to : g[v])
    if (to != p) {
      level[to] = 1 - level[v];
      dfs(to, v);
    }
  last[v] = b.size() - 1;
}
struct node {
  long long val, push;
  node(long long val = 0, long long push = 0) : val(val), push(push){};
};
vector<node> tree[2];
void push(long long v, long long l, long long r, vector<node>& tree) {
  long long m = (l + r) / 2;
  if (tree[v].push) {
    tree[v * 2].push += tree[v].push;
    tree[v * 2 + 1].push += tree[v].push;
    tree[v * 2].val += tree[v].push * (m - l + 1);
    tree[v * 2 + 1].val += tree[v].push * (r - m);
    tree[v].push = 0;
  }
}
void add(long long v, long long l, long long r, long long _l, long long _r,
         long long x, vector<node>& tree) {
  if (_l > _r) return;
  if (_r < l || _l > r) return;
  if (_l <= l && r <= _r) {
    tree[v].val += (r - l + 1) * x;
    tree[v].push += x;
    return;
  }
  push(v, l, r, tree);
  long long m = (l + r) / 2;
  add(v * 2, l, m, _l, _r, x, tree);
  add(v * 2 + 1, m + 1, r, _l, _r, x, tree);
  tree[v].val = tree[v * 2].val + tree[v * 2 + 1].val;
}
long long get(long long v, long long l, long long r, long long x,
              vector<node>& tree) {
  if (l > r) return 0;
  if (l == r) return tree[v].val;
  push(v, l, r, tree);
  long long m = (l + r) / 2;
  if (x <= m)
    return get(v * 2, l, m, x, tree);
  else
    return get(v * 2 + 1, m + 1, r, x, tree);
}
void build(long long v, long long l, long long r, vector<node>& tree,
           vector<long long>& k) {
  if (l == r) return void(tree[v].val = k[l]);
  long long m = (l + r) / 2;
  build(v * 2, l, m, tree, k);
  build(v * 2 + 1, m + 1, r, tree, k);
  tree[v].val = tree[v * 2].val + tree[v * 2 + 1].val;
}
signed main() {
  srand((long long)time(NULL));
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  vector<long long> a(n);
  for (auto& i : a) cin >> i;
  frst.resize(n), last.resize(n);
  g.resize(n);
  level.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  vector<long long> b1(n), b2(n);
  for (long long i = 0; i < n; i++)
    if (level[i])
      b1[frst[i]] = (a[i]), b2[frst[i]] = 0;
    else
      b2[frst[i]] = (a[i]), b1[frst[i]] = 0;
  tree[0].resize(4 * n);
  tree[1].resize(4 * n);
  build(1, 0, n - 1, tree[1], b1);
  build(1, 0, n - 1, tree[0], b2);
  while (m--) {
    long long tp;
    cin >> tp;
    if (tp == 1) {
      long long v, val;
      cin >> v >> val;
      v--;
      add(1, 0, n - 1, frst[v], last[v], val, tree[level[v]]);
      add(1, 0, n - 1, frst[v], last[v], -val, tree[1 - level[v]]);
    } else {
      long long v;
      cin >> v;
      v--;
      cout << get(1, 0, n - 1, frst[v], tree[level[v]]) << '\n';
    }
  }
}
