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>
long long INF = (1ll << 61);
using namespace std;
vector<long long> *g;
int disc[200005];
int st[200005];
int don[200005];
int a[200005];
int dep[200005];
int seqsz = 0;
void dfs(int V, int pV) {
  vector<long long> &adj = g[V];
  st[V] = seqsz;
  seqsz++;
  for (auto k : adj) {
    if (pV == k) continue;
    dep[k] = dep[V] ^ 1;
    dfs(k, V);
  }
  don[V] = seqsz - 1;
}
bool comp(int i, int j) { return st[i] < st[j]; }
template <class V, class U>
struct SegTree {
  int n;
  vector<V> st;
  vector<U> lazy;
  vector<int> leaves;
  SegTree(const int n) : n(n) {
    leaves.resize(n);
    init(1, 0, n - 1);
    lazy.resize(st.size());
  }
  void init(const int si, const int lo, const int hi) {
    if (lo == hi) {
      if (si >= (int)st.size()) st.resize(si + 1);
      st[si] = V();
      leaves[lo] = si;
    } else {
      const int mid = (lo + hi) >> 1;
      init(si << 1, lo, mid);
      init(si << 1 | 1, mid + 1, hi);
      st[si] = V(st[si << 1], st[si << 1 | 1]);
    }
  }
  void updateLazy(const int si, const int lo, const int hi) {
    lazy[si].apply(st[si], lo, hi);
    if (lo != hi) {
      lazy[si << 1] = U(lazy[si << 1], lazy[si]);
      lazy[si << 1 | 1] = U(lazy[si << 1 | 1], lazy[si]);
    }
    lazy[si] = U();
  }
  V query(const int l, const int r) {
    return (l <= r && l < n && r >= 0) ? query(l, r, 1, 0, n - 1) : V();
  }
  V query(const int l, const int r, const int si, const int lo, const int hi) {
    updateLazy(si, lo, hi);
    if (l <= lo && hi <= r) return st[si];
    const int mid = (lo + hi) >> 1;
    if (r <= mid) return query(l, r, si << 1, lo, mid);
    if (mid < l) return query(l, r, si << 1 | 1, mid + 1, hi);
    return V(query(l, r, si << 1, lo, mid),
             query(l, r, si << 1 | 1, mid + 1, hi));
  }
  void update(const int l, const int r, const U u) {
    if (l <= r) update(l, r, u, 1, 0, n - 1);
  }
  void update(const int l, const int r, const U &u, const int si, const int lo,
              const int hi) {
    if (l <= lo && hi <= r) {
      lazy[si] = U(lazy[si], u);
      updateLazy(si, lo, hi);
    } else {
      updateLazy(si, lo, hi);
      if (hi < l || r < lo) return;
      const int mid = (lo + hi) >> 1;
      update(l, r, u, si << 1, lo, mid);
      update(l, r, u, si << 1 | 1, mid + 1, hi);
      st[si] = V(st[si << 1], st[si << 1 | 1]);
    }
  }
};
struct StVal {
  long long v = 0;
  StVal() {}
  StVal(const long long v) : v(v) {}
  StVal(const StVal &v1, const StVal &v2) { v = v1 + v2; }
  operator long long() const { return v; }
};
struct StUpdate {
  long long v = 0;
  StUpdate() {}
  StUpdate(const long long v) : v(v) {}
  StUpdate(const StUpdate &u1, const StUpdate &u2) { v = u1 + u2; }
  operator long long() const { return v; }
  void apply(StVal &v, const int lo, const int hi) {
    v.v += this->v * (hi - lo + 1);
  }
};
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  g = new vector<long long>[n + 1];
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < n + 1; i++) disc[i] = i;
  dfs(1, -1);
  sort(disc + 1, disc + n + 1, comp);
  SegTree<StVal, StUpdate> tr[] = {SegTree<StVal, StUpdate>(n + 1),
                                   SegTree<StVal, StUpdate>(n + 1)};
  for (int i = 1; i < n + 1; i++) {
    tr[dep[i]].update(st[i], st[i], a[i]);
  }
  while (m--) {
    int t, a, b;
    cin >> t >> a;
    if (t == 1) {
      cin >> b;
      tr[dep[a]].update(st[a], don[a], b);
      tr[!dep[a]].update(st[a], don[a], -b);
    } else {
      cout << tr[dep[a]].query(st[a], st[a]) << endl;
    }
  }
  return 0;
}
