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>
#pragma GCC optimize("O3")
using namespace std;
const long long int mod = 1000000007;
const long long int MAX = 100008;
vector<long long int> v[200005];
long long int in[200005];
long long int out[200005];
long long int arr[200005];
long long int value[200005];
long long int level[200005];
int vis[200005];
long long int timer = 0;
struct node {
  long long int odd, even;
};
void dfs(long long int node) {
  vis[node] = 1;
  timer++;
  in[node] = timer;
  arr[timer] = node;
  for (int i = 0; i < v[node].size(); i++) {
    if (vis[v[node][i]] == 0) {
      level[v[node][i]] = level[node] + 1;
      dfs(v[node][i]);
    }
  }
  out[node] = timer;
}
struct node lazy[900009];
void initialize() {
  for (int i = 0; i < 900009; i++) {
    lazy[i].odd = 0;
    lazy[i].even = 0;
  }
}
long long int ans = 0;
void query(long long int start, long long int end, long long int node,
           long long int idx, long long int lev) {
  long long int mid = (start + end) >> 1;
  if (lazy[node].even != 0 && lazy[node].odd != 0 && start != end) {
    lazy[2 * node].even += lazy[node].even;
    lazy[2 * node + 1].even += lazy[node].even;
    lazy[2 * node].odd += lazy[node].odd;
    lazy[2 * node + 1].odd += lazy[node].odd;
    lazy[node].even = 0;
    lazy[node].odd = 0;
  }
  if (start > idx || idx > end) {
    return;
  } else if (start == end && idx == end) {
    if (lev & 1)
      ans += lazy[node].odd;
    else
      ans += lazy[node].even;
  } else {
    query(start, mid, 2 * node, idx, lev);
    query(mid + 1, end, (2 * node) + 1, idx, lev);
  }
}
void update(long long int start, long long int end, long long int node,
            long long int l, long long int r, long long int val,
            long long int lev) {
  long long int mid = (start + end) >> 1;
  if (lazy[node].even != 0 && lazy[node].odd != 0 && start != end) {
    lazy[2 * node].even += lazy[node].even;
    lazy[2 * node + 1].even += lazy[node].even;
    lazy[2 * node].odd += lazy[node].odd;
    lazy[2 * node + 1].odd += lazy[node].odd;
    lazy[node].even = 0;
    lazy[node].odd = 0;
  }
  if (start > r || l > end) {
    return;
  } else if (l <= start && r >= end) {
    if (lev % 2 == 0) {
      lazy[node].even += val;
      lazy[node].odd -= val;
    } else {
      lazy[node].even -= val;
      lazy[node].odd += val;
    }
    return;
  } else {
    update(start, mid, 2 * node, l, r, val, lev);
    update(mid + 1, end, (2 * node) + 1, l, r, val, lev);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> value[i];
  for (int i = 1; i <= n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  level[1] = 1;
  dfs(1);
  initialize();
  for (int i = 1; i <= n; i++) arr[in[i]] = value[i];
  while (q--) {
    int ch;
    cin >> ch;
    if (ch == 1) {
      long long int node, x;
      cin >> node >> x;
      long long int idx = in[node];
      update(1LL, n, 1LL, in[node], out[node], x, level[node]);
    } else {
      long long int node;
      cin >> node;
      ans = value[node];
      query(1, n, 1, in[node], level[node]);
      cout << ans << '\n';
    }
  }
}
