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 n, m, a[200001], x, y, z, second[200001], e[200001], k, st[1600001],
    lz[1600001], h[200001];
vector<int> v[200001];
void dfs(int x, int p) {
  second[x] = ++k;
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] == p) continue;
    h[v[x][i]] = h[x] + 1;
    dfs(v[x][i], x);
  }
  e[x] = ++k;
}
void push(int v) {
  st[v * 2] += lz[v];
  st[v * 2 + 1] += lz[v];
  lz[v * 2] += lz[v];
  lz[v * 2 + 1] += lz[v];
  lz[v] = 0;
}
void upd(int v, int A, int B, int l, int r, int x) {
  if (l > B || r < A) return;
  if (A == l && B == r) {
    st[v] += x;
    lz[v] += x;
    return;
  }
  push(v);
  int mid = (A + B) / 2;
  upd(v * 2, A, mid, l, min(mid, r), x);
  upd(v * 2 + 1, mid + 1, B, max(mid + 1, l), r, x);
}
int qr(int v, int A, int B, int x) {
  if (A == B) return st[v];
  push(v);
  int mid = (A + B) / 2;
  if (x <= mid)
    return qr(v * 2, A, mid, x);
  else
    return qr(v * 2 + 1, mid + 1, B, x);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++)
    cin >> x >> y, v[x].push_back(y), v[y].push_back(x);
  dfs(1, 0);
  while (m--) {
    cin >> x >> y;
    if (x == 1) {
      cin >> z;
      if (h[y] % 2)
        upd(1, 1, k, second[y], e[y], -z);
      else
        upd(1, 1, k, second[y], e[y], z);
    } else if (h[y] % 2)
      cout << a[y] - qr(1, 1, k, second[y]) << endl;
    else
      cout << a[y] + qr(1, 1, k, second[y]) << endl;
  }
  return 0;
}
