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>
const int maxn = 200000 + 10;
using namespace std;
struct Node {
  int start, end;
  int val;
  int d;
} node[maxn];
vector<int> g[maxn];
int n, m, cnt;
int c[2][maxn];
void dfs(int u, int fa, int d) {
  node[u].start = ++cnt;
  node[u].d = d;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u, 1 - d);
  }
  node[u].end = cnt;
}
void update(int *c, int idx, int val) {
  while (idx <= n) {
    c[idx] += val;
    idx += idx & -idx;
  }
}
int getsum(int *c, int idx) {
  int sum = 0;
  while (idx > 0) {
    sum += c[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int main() {
  int x, y, z, u, v;
  scanf("%d %d", &n, &m);
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= n; i++) scanf("%d", &node[i].val);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cnt = 0;
  dfs(1, -1, 0);
  while (m--) {
    scanf("%d", &x);
    if (x == 1) {
      scanf("%d %d", &y, &z);
      update(c[node[y].d], node[y].start, z);
      update(c[node[y].d], node[y].end + 1, -z);
    } else {
      scanf("%d", &y);
      int ans = node[y].val + getsum(c[node[y].d], node[y].start) -
                getsum(c[1 - node[y].d], node[y].start);
      cout << ans << endl;
    }
  }
  return 0;
}
