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 MAX_N = 200000 + 200;
pair<int, int> v[4 * MAX_N];
pair<int, int> flag[4 * MAX_N];
vector<int> adj[MAX_N];
int arr[MAX_N], MIN[4 * MAX_N], m, n, ans, s[MAX_N], e[MAX_N], h[MAX_N], t;
bool mark[MAX_N];
void dfs(int u) {
  mark[u] = true;
  s[u] = t;
  for (int i = 0; i < adj[u].size(); i++) {
    int w = adj[u][i];
    if (!mark[w]) {
      h[w] = h[u] + 1;
      dfs(w);
    }
  }
  e[u] = t;
  t++;
}
void make(int node, int l, int r) {
  v[node].first = l;
  v[node].second = r;
  if (l == r) return;
  int mid = (r + l) / 2;
  make(2 * node, l, mid);
  make(2 * node + 1, mid + 1, r);
}
void change(int node, int l, int r, int st, int en, int h, int val) {
  if (l == st && r == en) {
    if (h % 2 == 0)
      flag[node].first += val;
    else
      flag[node].second += val;
    return;
  }
  int mid = (r + l) / 2;
  if (en <= mid) {
    change(2 * node, l, mid, st, en, h, val);
    return;
  }
  if (mid < st) {
    change(2 * node + 1, mid + 1, r, st, en, h, val);
    return;
  }
  change(2 * node, l, mid, st, mid, h, val);
  change(2 * node + 1, mid + 1, r, mid + 1, en, h, val);
}
void FIND(int node, int l, int r, int tar) {
  if (l == r && r == e[tar]) {
    if (h[tar] % 2 == 0) {
      ans += flag[node].first;
      ans -= flag[node].second;
    } else {
      ans -= flag[node].first;
      ans += flag[node].second;
    }
    return;
  }
  int mid = (r + l) / 2;
  if (e[tar] <= mid) {
    if (h[tar] % 2 == 0) {
      ans += flag[node].first;
      ans -= flag[node].second;
    } else {
      ans -= flag[node].first;
      ans += flag[node].second;
    }
    FIND(2 * node, l, mid, tar);
    return;
  } else {
    if (h[tar] % 2 == 0) {
      ans += flag[node].first;
      ans -= flag[node].second;
    } else {
      ans -= flag[node].first;
      ans += flag[node].second;
    }
    FIND(2 * node + 1, mid + 1, r, tar);
    return;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  t = 1;
  dfs(1);
  make(1, 1, n);
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    if (a == 1) {
      int b, c;
      cin >> b >> c;
      change(1, 1, n, s[b], e[b], h[b], c);
    } else {
      int b;
      ans = 0;
      cin >> b;
      FIND(1, 1, n, b);
      cout << ans + arr[b] << endl;
    }
  }
}
