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;
void pv(vector<long long int> arr, long long int size) {
  for (long long int i = 0; i < size; i++) {
    cout << arr[i] << ' ';
  }
}
void gv(vector<long long int> &arr) {
  long long int size = arr.size();
  for (long long int i = 0; i < size; i++) {
    cin >> arr[i];
  }
}
long long int tim = 0;
void dfs(vector<vector<long long int>> &adj, long long int n,
         long long int vert, vector<long long int> &vis,
         vector<pair<long long int, long long int>> &a,
         vector<long long int> &level, long long int ht) {
  vis[vert] = 1;
  level[vert] = ht;
  a[vert].first = tim++;
  for (long long int i = 0; i < adj[vert].size(); i++) {
    if (vis[adj[vert][i]] == 0)
      dfs(adj, n, adj[vert][i], vis, a, level, ht + 1);
  }
  a[vert].second = tim++;
}
class segTree {
  long long int identity, size;
  vector<long long int> segtree, arr;
  long long int f(long long int x, long long int y) { return x + y; }
  void buildseg() {
    long long int n = size;
    for (long long int i = n - 1; i > 0; --i)
      segtree[i] = f(segtree[i << 1], segtree[i << 1 | 1]);
  }
  void update(long long int p, long long int value) {
    long long int n = size;
    for (segtree[p += n] = value; p > 1; p >>= 1)
      segtree[p >> 1] = f(segtree[p], segtree[p ^ 1]);
  }
  long long int query(long long int l, long long int r) {
    long long int resl = identity, resr = identity, n = size;
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) resl = f(resl, segtree[l++]);
      if (r & 1) resr = f(segtree[--r], resr);
    }
    return f(resl, resr);
  }
  void buildsegrec(long long int start, long long int end, long long int ind) {
    if (start == end)
      segtree[ind] = arr[start];
    else {
      long long int mid = (end - start) / 2 + start;
      buildsegrec(start, mid, 2 * ind + 1);
      buildsegrec(mid + 1, end, 2 * ind + 2);
      segtree[ind] = f(segtree[2 * ind + 1], segtree[2 * ind + 2]);
    }
  }
  void updaterec(long long int ind, long long int i_to_update,
                 long long int val, long long int start, long long int end) {
    if (start == end) {
      arr[i_to_update] += val;
      segtree[ind] += val;
      return;
    }
    long long int mid = (end - start) / 2 + start;
    if (i_to_update > mid) {
      updaterec(2 * ind + 2, i_to_update, val, mid + 1, end);
    } else {
      updaterec(2 * ind + 1, i_to_update, val, start, mid);
    }
    segtree[ind] = f(segtree[2 * ind + 1], segtree[2 * ind + 2]);
  }
  long long int queryrec(long long int ind, long long int start,
                         long long int end, long long int left,
                         long long int right) {
    if (right < start || left > end) return identity;
    if (left <= start && right >= end) {
      return segtree[ind];
    }
    long long int mid = (end - start) / 2 + start;
    long long int q1 = queryrec(2 * ind + 1, start, mid, left, right);
    long long int q2 = queryrec(2 * ind + 2, mid + 1, end, left, right);
    return f(q1, q2);
  }

 public:
  segTree(long long int n, vector<long long int> &ar, long long int ident,
          long long int def) {
    segtree = vector<long long int>(4 * n, def);
    arr = ar;
    identity = ident;
    buildsegrec(0, n - 1, 0);
    size = n;
  }
  void updateTree(long long int i_to_update, long long int val) {
    updaterec(0, i_to_update, val, 0, size - 1);
  }
  long long int querySeg(long long int l, long long int r) {
    return queryrec(0, 0, size - 1, l, r);
  }
  void printSeg() { pv(segtree, 4 * size); }
};
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m;
  cin >> n >> m;
  vector<long long int> arr(n);
  vector<vector<long long int>> adj(n + 1);
  gv(arr);
  for (long long int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  vector<pair<long long int, long long int>> a(n + 1);
  vector<long long int> vis(n + 1, 0);
  vector<long long int> dd(2 * n, 0), level(n + 1, 0);
  vector<long long int> pp(2 * n, 0);
  dfs(adj, n, 1, vis, a, level, 0);
  segTree s(2 * n, dd, 0, 0);
  for (long long int i = 0; i < m; i++) {
    long long int x, y, z;
    cin >> x;
    if (x == 1) {
      cin >> y >> z;
      if (level[y] % 2 == 1) {
        s.updateTree(a[y].first, +z);
        if (a[y].second + 1 < 2 * n) s.updateTree(a[y].second + 1, -z);
      } else {
        s.updateTree(a[y].first, -z);
        if (a[y].second + 1 < 2 * n) s.updateTree(a[y].second + 1, z);
      }
    } else {
      cin >> y;
      long long int sum = s.querySeg(0, a[y].first);
      if (level[y] % 2 == 1)
        cout << arr[y - 1] + sum << '\n';
      else
        cout << arr[y - 1] - sum << '\n';
    }
  }
  return 0;
}
