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 N = 400000 + 10;
#pragma comment(linker, "/STACK:102400000,102400000")
struct Edge {
  int v, nex;
  Edge() {}
  Edge(int _v, int _nex) {
    v = _v;
    nex = _nex;
  }
} eg[N];
int g[N], idx;
int id[N], son[N];
int a[N], dep;
int pre[N];
int kkcld[N];
void addedge(int u, int v) {
  eg[idx] = Edge(v, g[u]);
  g[u] = idx++;
}
void dfs(int u, int fa, int v) {
  son[u] = 1;
  a[++dep] = u;
  id[u] = dep;
  kkcld[u] = v;
  for (int i = g[u]; ~i; i = eg[i].nex) {
    int to = eg[i].v;
    if (to == fa) continue;
    dfs(to, u, -v);
    son[u] += son[to];
  }
}
struct node {
  int sum, addv;
};
struct Seg {
  node tree[N << 2];
  void Up(node &rt, node &ls, node &rs) { rt.sum = ls.sum + rs.sum; }
  void Down(node &rt, node &ls, node &rs, int m) {
    if (rt.addv) {
      ls.sum += (m - (m >> 1)) * rt.addv;
      rs.sum += (m >> 1) * rt.addv;
      ls.addv += rt.addv, rs.addv += rt.addv;
      rt.addv = 0;
    }
  }
  void build(int l, int r, int rt) {
    tree[rt].addv = 0;
    if (l == r) {
      tree[rt].sum = pre[a[l]];
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    Up(tree[rt], tree[rt << 1], tree[rt << 1 | 1]);
  }
  void update_add(int L, int R, int v, int l, int r, int rt) {
    if (L == l && R == r) {
      tree[rt].addv += v;
      tree[rt].sum += (R - L + 1) * v;
      return;
    }
    int mid = (l + r) >> 1;
    Down(tree[rt], tree[rt << 1], tree[rt << 1 | 1], r - l + 1);
    if (R <= mid)
      update_add(L, R, v, l, mid, rt << 1);
    else if (L > mid)
      update_add(L, R, v, mid + 1, r, rt << 1 | 1);
    else {
      update_add(L, mid, v, l, mid, rt << 1);
      update_add(mid + 1, R, v, mid + 1, r, rt << 1 | 1);
    }
    Up(tree[rt], tree[rt << 1], tree[rt << 1 | 1]);
  }
  int query_sum(int L, int R, int l, int r, int rt) {
    if (L == l && R == r) {
      return tree[rt].sum;
    }
    int mid = (l + r) >> 1;
    Down(tree[rt], tree[rt << 1], tree[rt << 1 | 1], r - l + 1);
    if (L > mid)
      return query_sum(L, R, mid + 1, r, rt << 1 | 1);
    else if (R <= mid)
      return query_sum(L, R, l, mid, rt << 1);
    else
      return query_sum(L, mid, l, mid, rt << 1) +
             query_sum(mid + 1, R, mid + 1, r, rt << 1 | 1);
  }
} sega, segb;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &pre[i]);
  int u, v;
  idx = 0;
  memset(g, -1, sizeof g);
  for (int i = 2; i <= n; ++i) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  dep = 0;
  dfs(1, -1, 1);
  sega.build(1, dep, 1);
  segb.build(1, dep, 1);
  int typ;
  while (m-- > 0) {
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%d%d", &u, &v);
      if (kkcld[u] == 1) {
        sega.update_add(id[u], id[u] + son[u] - 1, v, 1, dep, 1);
        segb.update_add(id[u], id[u] + son[u] - 1, -v, 1, dep, 1);
      } else {
        segb.update_add(id[u], id[u] + son[u] - 1, v, 1, dep, 1);
        sega.update_add(id[u], id[u] + son[u] - 1, -v, 1, dep, 1);
      }
    } else {
      scanf("%d", &u);
      if (kkcld[u] == 1)
        printf("%d\n", sega.query_sum(id[u], id[u], 1, dep, 1));
      else
        printf("%d\n", segb.query_sum(id[u], id[u], 1, dep, 1));
    }
  }
  return 0;
}
