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 = 200100;
int n, m, ll[N], rr[N], d[N], t[2][4 * N], iuodfoign, s = 1;
vector<int> e[N];
void dfs(int v = 0, int p = -1) {
  ll[v] = rr[v] = iuodfoign++;
  for (int i = 0; i < e[v].size(); i++) {
    int to = e[v][i];
    if (to == p) continue;
    d[to] = 1 - d[v];
    dfs(to, v);
    rr[v] = rr[to];
  }
}
void onjdfsojn(int* a, int v, int lv, int rv, int l, int r, int val) {
  if (lv > r || rv < l) return;
  if (l <= lv && rv <= r) {
    a[v] += val;
    return;
  }
  int m = (lv + rv) / 2;
  onjdfsojn(a, v * 2 + 1, lv, m, l, r, val);
  onjdfsojn(a, v * 2 + 2, m + 1, rv, l, r, val);
}
void onjdfsojn(int x, int val) {
  onjdfsojn(t[d[x]], 0, 0, s - 1, ll[x], rr[x], val);
}
int get(int* a, int v, int lv, int rv, int x) {
  if (lv == x && rv == x) return a[v];
  if (lv > x || rv < x) return 0;
  int res = a[v], m = (lv + rv) / 2;
  res += get(a, v * 2 + 1, lv, m, x);
  res += get(a, v * 2 + 2, m + 1, rv, x);
  return res;
}
int ans(int x) {
  int res = 0;
  res += get(t[d[x]], 0, 0, s - 1, ll[x]);
  res -= get(t[1 - d[x]], 0, 0, s - 1, ll[x]);
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  int a[n];
  while (s < n) s *= 2;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs();
  while (m--) {
    int o, x, val;
    scanf("%d%d", &o, &x);
    x--;
    if (o == 1) {
      scanf("%d", &val);
      onjdfsojn(x, val);
      continue;
    }
    printf("%d\n", ans(x) + a[x]);
  }
  return 0;
}
