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 = 2000 * 100 + 10;
int n, m, a[maxn], s[maxn], f[maxn], cnt = 1, h[maxn], sum1[4 * maxn],
                                     sum0[4 * maxn], ja[maxn];
vector<int> adj[maxn];
bool mark[maxn];
vector<int> st;
int chert[maxn];
void dfs(int x) {
  mark[x] = true;
  s[x] = cnt++;
  chert[st.size()] = x;
  st.push_back(s[x]);
  for (int i = 0; i < (int)adj[x].size(); i++) {
    int next = adj[x][i];
    if (!mark[next]) {
      h[next] = h[x] + 1;
      dfs(next);
    }
  }
  f[x] = cnt++;
}
void change(int x, int y, int carr, int hal, int lo = 0, int hi = n - 1,
            int cnt = 1) {
  if (lo > y || hi < x) return;
  if (lo >= x && hi <= y) {
    if (hal) {
      sum1[cnt] += carr;
      sum0[cnt] -= carr;
    } else {
      sum1[cnt] -= carr;
      sum0[cnt] += carr;
    }
    return;
  }
  int mid = (lo + hi) / 2;
  change(x, y, carr, hal, lo, mid, cnt * 2);
  change(x, y, carr, hal, mid + 1, hi, cnt * 2 + 1);
}
int find(int x, int hal, int lo = 0, int hi = n - 1, int cnt = 1) {
  if (lo > x || hi < x) return 0;
  if (lo == hi) {
    int total = a[chert[x]];
    if (hal)
      total += sum1[cnt];
    else
      total += sum0[cnt];
    return total;
  }
  int mid = (lo + hi) / 2;
  if (hal)
    return find(x, hal, lo, mid, cnt * 2) +
           find(x, hal, mid + 1, hi, cnt * 2 + 1) + sum1[cnt];
  else
    return find(x, hal, lo, mid, cnt * 2) +
           find(x, hal, mid + 1, hi, cnt * 2 + 1) + sum0[cnt];
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0);
  for (int i = 0; i < n; i++) ja[chert[i]] = i;
  for (int i = 0; i < m; i++) {
    int hw, x, val;
    cin >> hw;
    if (hw == 1) {
      cin >> x >> val;
      x--;
      int q1 = lower_bound(st.begin(), st.end(), s[x]) - st.begin();
      int q2 = lower_bound(st.begin(), st.end(), f[x]) - st.begin();
      q2--;
      change(q1, q2, val, h[x] % 2);
    } else {
      cin >> x;
      x--;
      cout << find(ja[x], h[x] % 2) << endl;
    }
  }
  return 0;
}
