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;
int lg(long long n) {
  assert(n > 0);
  int ans = -1;
  while (n) {
    ans++;
    n >>= 1;
  }
  return ans;
}
struct SegmentTree {
  int n;
  vector<long long> mn, add;
  SegmentTree(int n) : n(n), mn(4 * n, 0), add(4 * n) {}
  void Add(int id, int l, int r, long long val) {
    add[id] += val;
    mn[id] += val;
  }
  void push(int id, int l, int r) {
    if (add[id] != 0) {
      int m = (l + r) >> 1;
      Add(id << 1, l, m, add[id]);
      Add(id << 1 | 1, m + 1, r, add[id]);
      add[id] = 0;
    }
  }
  void pull(int id) { mn[id] = min(mn[id << 1], mn[id << 1 | 1]); }
  SegmentTree(const vector<int> &v) {
    n = (int)v.size();
    mn.resize(4 * n);
    add.resize(4 * n);
    function<void(int, int, int)> build = [&](int id, int l, int r) {
      if (l == r) {
        mn[id] = v[l];
        return;
      }
      int m = (l + r) >> 1;
      build(id << 1, l, m);
      build(id << 1 | 1, m + 1, r);
      pull(id);
    };
    build(1, 0, n - 1);
  }
  void RangeAdd(int id, int l, int r, int ql, int qr, long long val) {
    if (l > qr || r < ql) return;
    if (ql <= l && qr >= r) {
      Add(id, l, r, val);
      return;
    }
    push(id, l, r);
    int m = (l + r) >> 1;
    RangeAdd(id << 1, l, m, ql, qr, val);
    RangeAdd(id << 1 | 1, m + 1, r, ql, qr, val);
    pull(id);
  }
  void RangeAdd(int ql, int qr, long long val) {
    RangeAdd(1, 0, n - 1, ql, qr, val);
  }
  long long GetMin(int id, int l, int r, int ql, int qr) {
    if (l > qr || r < ql) return 9e18;
    if (ql <= l && qr >= r) return mn[id];
    push(id, l, r);
    int m = (l + r) >> 1;
    return min(GetMin(id << 1, l, m, ql, qr),
               GetMin(id << 1 | 1, m + 1, r, ql, qr));
  }
  long long GetMin(int ql, int qr) { return GetMin(1, 0, n - 1, ql, qr); }
  long long GetEle(int pos) { return GetMin(1, 0, n - 1, pos, pos); }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<int>> e(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  int T = 0;
  vector<int> tin(n), tout(n), h(n);
  function<void(int, int)> dfs = [&](int cur, int fa) {
    tin[cur] = tout[cur] = T++;
    for (int nxt : e[cur]) {
      if (nxt == fa) continue;
      h[nxt] = h[cur] + 1;
      dfs(nxt, cur);
      tout[cur] = max(tout[nxt], tout[cur]);
    }
  };
  h[0] = 0;
  dfs(0, -1);
  SegmentTree st(n);
  while (m--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, val;
      cin >> x >> val;
      x--;
      int sign = (h[x] & 1 ? -1 : 1);
      st.RangeAdd(tin[x], tout[x], sign * val);
    } else if (op == 2) {
      int x;
      cin >> x;
      x--;
      int sign = (h[x] & 1 ? -1 : 1);
      cout << a[x] + sign * st.GetEle(tin[x]) << '\n';
    } else
      assert(false);
  }
  return 0;
}
