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 <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
const int infi = 1e9;
const long long Linfi = (long long)1e18;
const long long MOD = 1000000007;
const int c1 = 31;
int n, m, u, v, cnt = 0;
vector<int> adj[400005];
int a[400005], depth[400005], in[400005], out[400005], xet[400005];
int bit[400005 + 10];
void update(int first, int val) {
  while (first < 400005) {
    bit[first] += val;
    first += (first & (-first));
  }
}
int get(int first) {
  int ans = 0;
  while (first) {
    ans += bit[first];
    first -= (first & (-first));
  }
  return ans;
}
void DFS(int u) {
  xet[u] = 1;
  in[u] = ++cnt;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (xet[v] == 0) {
      depth[v] = depth[u] + 1;
      DFS(v);
    }
  }
  out[u] = ++cnt;
}
void solve() {
  DFS(1);
  while (m--) {
    int cmd, first, val;
    cin >> cmd;
    if (cmd == 1) {
      cin >> first >> val;
      if (depth[first] % 2 == 0) {
        update(in[first], val);
        update(out[first], -val);
      } else {
        update(in[first], -val);
        update(out[first], val);
      }
    } else {
      cin >> first;
      if (depth[first] % 2 == 0) {
        cout << a[first] + get(in[first]) << endl;
      } else {
        cout << a[first] - get(in[first]) << endl;
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  solve();
  return 0;
}
