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 N = 200100;
vector<int> graf[N];
int cost[N];
int n, q;
inline void load() {
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &cost[i]);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    --x;
    --y;
    graf[x].push_back(y);
    graf[y].push_back(x);
  }
}
int disc[N], fin[N], depth[N];
int timer = 0;
void dfs(int x, int otac = -1, int dub = 0) {
  depth[x] = dub;
  disc[x] = ++timer;
  for (int i = 0; i < graf[x].size(); i++)
    if (graf[x][i] != otac) dfs(graf[x][i], x, dub + 1);
  fin[x] = timer;
}
struct fenwick {
  int tree[200100];
  int n;
  fenwick() {
    memset(tree, 0, sizeof(tree));
    n = 200010;
  }
  inline void add(int x, int val) {
    if (x == 0) return;
    for (; x < n; x += x & -x) tree[x] += val;
  }
  inline int get(int x) {
    int ans = 0;
    for (; x; x -= x & -x) ans += tree[x];
    return ans;
  }
} F[2];
inline void solve() {
  dfs(0);
  for (; q; q--) {
    int ty;
    scanf("%d", &ty);
    if (ty == 2) {
      int node;
      scanf("%d", &node);
      node--;
      int dub = depth[node];
      printf("%d\n", F[dub % 2].get(disc[node]) + cost[node]);
    } else {
      int node, val;
      scanf("%d %d", &node, &val);
      node--;
      int dub = depth[node];
      F[dub % 2].add(fin[node] + 1, -val);
      F[dub % 2].add(disc[node], val);
      F[!(dub % 2)].add(fin[node] + 1, val);
      F[!(dub % 2)].add(disc[node], -val);
    }
  }
}
int main(void) {
  load();
  solve();
  return 0;
}
