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;
const int MAXN = 200111;
int wej[MAXN], subts[MAXN];
vector<int> kraw[MAXN];
int ojc[MAXN];
int dep[MAXN];
int nrwe = 1;
void dfs(int v) {
  wej[v] = nrwe++;
  subts[v] = 1;
  for (auto b : kraw[v]) {
    if (b == ojc[v]) continue;
    ojc[b] = v;
    dep[b] = dep[v] + 1;
    dfs(b);
    subts[v] += subts[b];
  }
}
const int SS = 1 << 19;
int tree[2 * SS + 6969];
int st, en, val;
void insert(int a, int b, int v) {
  if (st <= a && b <= en) {
    tree[v] += val;
    return;
  }
  int sr = (a + b) / 2;
  if (st <= sr) insert(a, sr, 2 * v);
  if (sr < en) insert(sr + 1, b, 2 * v + 1);
}
int pytaj(int v) {
  v += SS - 1;
  int wyn = 0;
  while (v > 0) wyn += tree[v], v /= 2;
  return wyn;
}
int tab[MAXN];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &tab[i]);
  for (int i = 2; i <= n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    kraw[a].push_back(b);
    kraw[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    if (dep[i] % 2) tab[i] = -tab[i];
    tree[wej[i] + SS - 1] = tab[i];
  }
  while (m--) {
    int typ;
    scanf("%d", &typ);
    if (typ == 1) {
      int a, vl;
      scanf("%d%d", &a, &vl);
      if (dep[a] % 2) vl = -vl;
      val = vl;
      st = wej[a], en = wej[a] + subts[a] - 1;
      insert(1, SS, 1);
    } else {
      int x;
      scanf("%d", &x);
      int ans = pytaj(wej[x]);
      if (dep[x] % 2) ans = -ans;
      printf("%d\n", ans);
    }
  }
}
