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 N = 2e5 + 2;
const int M = 1e6 + 2;
const int mod = 998244353;
const int inf = 2e9 + 3;
const long long INF = 1e16;
int n, m, a[N], pos[N], tick, sz[N], d[N];
vector<int> g[N];
struct tree {
  int t[M] = {0};
  void upd(int l, int r, int val, int v = 1, int tl = 1, int tr = n) {
    if (l <= tl && tr <= r) return void(t[v] += val);
    if (tr < l || r < tl) return;
    int tm = tl + tr >> 1;
    upd(l, r, val, v * 2, tl, tm);
    upd(l, r, val, v * 2 + 1, tm + 1, tr);
  }
  int get(int pos, int v = 1, int tl = 1, int tr = n) {
    if (tl == tr) return t[v];
    int tm = tl + tr >> 1;
    if (pos <= tm)
      return t[v] + get(pos, v * 2, tl, tm);
    else
      return t[v] + get(pos, v * 2 + 1, tm + 1, tr);
  }
} t[2];
void dfs(int v = 1, int p = 0, int c = 0) {
  pos[v] = ++tick;
  d[v] = c;
  sz[v] = 1;
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, v, !c);
    sz[v] += sz[to];
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1, v, u; i < n; ++i) {
    scanf("%d %d", &v, &u);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs();
  for (int i = 1; i <= n; ++i) t[d[i]].upd(pos[i], pos[i], a[i]);
  while (m--) {
    int T, v, x;
    scanf("%d %d", &T, &v);
    if (T == 2) {
      printf("%d\n", t[d[v]].get(pos[v]));
    } else {
      scanf("%d", &x);
      t[d[v]].upd(pos[v], pos[v] + sz[v] - 1, x);
      t[!d[v]].upd(pos[v], pos[v] + sz[v] - 1, -x);
    }
  }
}
