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;
struct BIT {
  vector<long long> bit;
  int n;
  BIT(int sz) {
    bit.resize(sz + 1);
    n = sz;
  }
  void add(int x, long long v) {
    while (x <= n) {
      bit[x] += v;
      x += x & (-x);
    }
  }
  void add(int l, int r, long long v) {
    if (l > r) return;
    add(l, v);
    add(r + 1, -v);
  }
  long long ask(int x) {
    long long ret = 0;
    while (x) {
      ret += bit[x];
      x -= x & (-x);
    }
    return ret;
  }
};
int arr[200005], l[200005], r[200005], dep[200005];
vector<int> G[200005];
int n, m, times = 0;
void dfs(int x, int p, int d) {
  l[x] = ++times;
  dep[x] = d;
  for (int y : G[x]) {
    if (y != p) dfs(y, x, d + 1);
  }
  r[x] = ++times;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", arr + i);
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, -1, 1);
  int op, x, val;
  BIT loli(2 * n), LOLI(2 * n);
  while (m--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &x, &val);
      if (dep[x] & 1) {
        LOLI.add(l[x], r[x], val);
        loli.add(l[x], r[x], -val);
      } else {
        loli.add(l[x], r[x], val);
        LOLI.add(l[x], r[x], -val);
      }
    } else {
      scanf("%d", &x);
      long long ans = arr[x];
      if (dep[x] & 1) {
        ans = ans + LOLI.ask(l[x]);
      } else {
        ans = ans + loli.ask(l[x]);
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}
