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;
long long n;
vector<long long> a(200005), seg(4 * 200005);
long long lazy[4 * 200005][2];
long long b[200005], lvl[200005], st_time[200005], end_time[200005], T = 0;
vector<long long> g[200005];
long long assign(long long a, long long b) { return a + b; }
void build(long long pos = 0, long long l = 0, long long r = n - 1) {
  if (l == r) {
    seg[pos] = b[a[l]];
    return;
  }
  long long mid = (l + r) / 2;
  build(2 * pos + 1, l, mid);
  build(2 * pos + 2, mid + 1, r);
  seg[pos] = assign(seg[2 * pos + 1], seg[2 * pos + 2]);
}
long long node_level(long long i) {
  long long x = a[i];
  return lvl[x];
}
void upd(long long pos, long long l, long long r, long long x, long long y) {
  lazy[pos][0] += x;
  lazy[pos][1] += y;
  if (l == r) {
    long long ll = node_level(l);
    if (ll)
      seg[pos] += y - x;
    else
      seg[pos] += x - y;
  }
}
void shift(long long pos, long long l, long long r) {
  long long mid = (l + r) / 2;
  upd(2 * pos + 1, l, mid, lazy[pos][0], lazy[pos][1]);
  upd(2 * pos + 2, mid + 1, r, lazy[pos][0], lazy[pos][1]);
  lazy[pos][0] = lazy[pos][1] = 0;
}
void update(long long x, long long y, long long v1, long long v2,
            long long pos = 0, long long l = 0, long long r = n - 1) {
  if (x > r || y < l) return;
  if (x <= l && y >= r) {
    upd(pos, l, r, v1, v2);
    return;
  }
  shift(pos, l, r);
  long long mid = (l + r) / 2;
  update(x, y, v1, v2, 2 * pos + 1, l, mid);
  update(x, y, v1, v2, 2 * pos + 2, mid + 1, r);
  seg[pos] = assign(seg[2 * pos + 1], seg[2 * pos + 2]);
}
long long query(long long x, long long y, long long pos = 0, long long l = 0,
                long long r = n - 1) {
  if (x > r || y < l) return 0;
  if (x <= l && y >= r) return seg[pos];
  shift(pos, l, r);
  long long mid = (l + r) / 2;
  return assign(query(x, y, 2 * pos + 1, l, mid),
                query(x, y, 2 * pos + 2, mid + 1, r));
}
void dfs(long long u, long long p) {
  lvl[u] = 1 - lvl[p];
  a[T] = u;
  st_time[u] = T++;
  for (auto v : g[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
  end_time[u] = T;
}
void solve() {
  long long m, i, x, y, z;
  cin >> n >> m;
  for (i = 0; i < n; i++) cin >> b[i];
  for (i = 1; i < n; i++) {
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  lvl[0] = 0;
  dfs(0, 0);
  build();
  while (m--) {
    cin >> x >> y;
    y--;
    if (x == 1) {
      cin >> z;
      long long l = st_time[y];
      long long r = end_time[y] - 1;
      long long level = lvl[y];
      if (level)
        update(l, r, 0, z);
      else
        update(l, r, z, 0);
    } else {
      long long val = st_time[y];
      cout << query(val, val) << "\n";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) solve();
  return 0;
}
