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 MAXINT = 1 << 31 - 1;
const long long MAXLONG = (long long)1 << 63 - 1;
const int MAXN = 2 * 1e5 + 10;
const int MAXN4 = 4 * MAXN;
int odd[MAXN4], eve[MAXN4], arr[MAXN], sta[MAXN], fin[MAXN], level[MAXN];
vector<int> gra[MAXN];
int n, m, tt;
void dfs(int u, int par, int lev) {
  level[u] = lev;
  sta[u] = tt;
  tt++;
  for (int i = (0); i < (gra[u].size()); i += 1) {
    int v = gra[u][i];
    if (v == par) continue;
    dfs(v, u, lev + 1);
  }
  fin[u] = tt;
  tt++;
}
void update(int arr[], int val, int x) {
  int cur = x;
  while (cur < MAXN4) {
    arr[cur] += val;
    cur += cur & (-cur);
  }
}
int query(int arr[], int x) {
  int cur = x;
  int res = 0;
  while (cur >= 1) {
    res += arr[cur];
    cur -= cur & (-cur);
  }
  return res;
}
int main() {
  while (scanf("%d %d", &n, &m) != -1) {
    for (int i = (1); i < (n + 1); i += 1) {
      scanf("%d", &arr[i]);
    }
    memset(odd, 0, sizeof(odd));
    memset(eve, 0, sizeof(eve));
    for (int i = (0); i < (n - 1); i += 1) {
      int x, y;
      scanf("%d %d", &x, &y);
      gra[x].push_back(y);
      gra[y].push_back(x);
    }
    tt = 1;
    dfs(1, -1, 1);
    for (int i = (0); i < (m); i += 1) {
      int type;
      scanf("%d", &type);
      if (type == 1) {
        int x, val;
        scanf("%d %d", &x, &val);
        if (level[x] % 2 == 0) {
          update(eve, val, sta[x]);
          update(eve, -val, fin[x] + 1);
          update(odd, -val, sta[x]);
          update(odd, val, fin[x] + 1);
        } else {
          update(odd, val, sta[x]);
          update(odd, -val, fin[x] + 1);
          update(eve, -val, sta[x]);
          update(eve, val, fin[x] + 1);
        }
      } else {
        int x, res;
        scanf("%d", &x);
        if (level[x] % 2 == 0) {
          res = query(eve, sta[x]);
        } else {
          res = query(odd, sta[x]);
        }
        res += arr[x];
        printf("%d\n", res);
      }
    }
  }
  return 0;
}
