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 mx = 2e5 + 5;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
int tree[2][mx << 2];
int lazy[2][mx << 2];
vector<int> adj[mx];
int cost[mx], arr[mx];
int dis[mx], fin[mx];
int color[mx];
int timer = 0;
void dfs(int u, int p, int c) {
  dis[u] = ++timer;
  color[u] = c;
  arr[timer] = cost[u];
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs(v, u, 1 - c);
  }
  fin[u] = timer;
}
void update(int id, int node, int b, int e, int l, int r, int val) {
  int left = node << 1;
  int right = left | 1;
  int mid = (b + e) >> 1;
  if (lazy[id][node] != 0) {
    tree[id][node] += (e - b + 1) * lazy[id][node];
    if (e != b) {
      lazy[id][left] += lazy[id][node];
      lazy[id][right] += lazy[id][node];
    }
    lazy[id][node] = 0;
  }
  if (b > r || l > e || b > e) return;
  if (b >= l && e <= r) {
    tree[id][node] += (e - b + 1) * val;
    if (e != b) {
      lazy[id][left] += val;
      lazy[id][right] += val;
    }
    return;
  }
  update(id, left, b, mid, l, r, val);
  update(id, right, mid + 1, e, l, r, val);
  tree[id][node] = tree[id][left] + tree[id][right];
}
int query(int id, int node, int b, int e, int l, int r) {
  int left = node << 1;
  int right = left | 1;
  int mid = (b + e) >> 1;
  if (lazy[id][node] != 0) {
    tree[id][node] += (e - b + 1) * lazy[id][node];
    if (e != b) {
      lazy[id][left] += lazy[id][node];
      lazy[id][right] += lazy[id][node];
    }
    lazy[id][node] = 0;
  }
  if (b > r || l > e) return 0;
  if (b >= l && e <= r) return tree[id][node];
  int x = query(id, left, b, mid, l, r);
  int y = query(id, right, mid + 1, e, l, r);
  return x + y;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &cost[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 1, 0);
  for (int i = 1; i <= n; i++) {
    update(0, 1, 1, n, i, i, arr[i]);
    update(1, 1, 1, n, i, i, arr[i]);
  }
  while (q--) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int x, val;
      scanf("%d%d", &x, &val);
      int c = color[x];
      int l = dis[x];
      int r = fin[x];
      update(c, 1, 1, n, l, r, val);
      update(1 - c, 1, 1, n, l, r, -val);
    } else {
      int x;
      scanf("%d", &x);
      int c = color[x];
      int ans = query(c, 1, 1, n, dis[x], dis[x]);
      printf("%d\n", ans);
    }
  }
  return 0;
}
