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;
struct Tree {
  int start;
  int end;
  set<int> child;
  int sign;
} t[200001];
int segadd[524288];
void update(int l, int r, int v, int id = 1, int cl = 0, int cr = 262144) {
  if (l <= cl and cr <= r) {
    segadd[id] += v;
    return;
  }
  if (r <= cl or l >= cr) return;
  int x = (cl + cr) / 2;
  update(l, r, v, id * 2 + 0, cl, x);
  update(l, r, v, id * 2 + 1, x, cr);
}
int get(int x) {
  x += 262144;
  int res = 0;
  while (x > 0) {
    res += segadd[x];
    x /= 2;
  }
  return res;
}
int vertices = 1;
void prepare(int x) {
  t[x].start = vertices++;
  for (auto y : t[x].child) {
    t[y].sign = -t[x].sign;
    t[y].child.erase(x);
    prepare(y);
  }
  t[x].end = vertices;
}
int main() {
  int n, m;
  cin >> n >> m;
  int a[n];
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int b, c;
    cin >> b >> c;
    t[b].child.insert(c);
    t[c].child.insert(b);
  }
  t[1].sign = +1;
  prepare(1);
  for (int i = 1; i <= n; i++) {
    update(t[i].start, t[i].start + 1, a[i] * t[i].sign);
  }
  for (int i = 0; i < m; i++) {
    int j, k, l;
    cin >> j >> k;
    if (j == 1) {
      cin >> l;
      update(t[k].start, t[k].end, l * t[k].sign);
    } else {
      cout << get(t[k].start) * t[k].sign << endl;
    }
  }
}
