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;
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
const int mx = 2e5 + 5;
const int INF = 1000000007;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y >>= 1;
    x = (x * x) % 1000000007;
  }
  return (res % 1000000007);
}
class BIT {
  vector<long long> bit;
  long long N;

 public:
  void update(long long x, long long val) {
    long long ind = x;
    while (ind <= N) {
      bit[ind] += val;
      ind += ((ind) & (-ind));
    }
  }
  BIT(long long n) {
    N = mx;
    bit = vector<long long>(N, 0);
  }
  long long getSum(long long x) {
    if (x < 0) return 0;
    long long ind = x;
    long long sum = 0;
    while (ind > 0) {
      sum += bit[ind];
      ind -= ((ind) & (-ind));
    }
    return sum;
  }
  long long getValue(long long x) { return getSum(x) - getSum(x - 1); }
  void changeElem(long long x, long long val) {
    update(x + 1, val - getValue(x));
  }
};
vector<long long> adj[mx];
vector<long long> in, out;
long long timer = 0;
vector<long long> depth;
void dfs(long long curr, long long par, long long d) {
  in[curr] = ++timer;
  depth[curr] = d;
  for (auto child : adj[curr]) {
    if (child != par) {
      dfs(child, curr, d + 1);
    }
  }
  out[curr] = timer;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  in = vector<long long>(n + 10);
  out = vector<long long>(n + 10);
  depth = vector<long long>(n + 10);
  vector<long long> arr(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0, 0);
  BIT bit(n);
  while (m--) {
    long long op, x;
    cin >> op >> x;
    if (op == 1) {
      long long val;
      cin >> val;
      if (depth[x] & 1) val = -val;
      bit.update(in[x], val);
      bit.update(out[x] + 1, -val);
    } else {
      long long extra = bit.getSum(in[x]);
      if (depth[x] & 1) extra = -extra;
      cout << extra + arr[x] << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}
