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 MAXN = 2e5 + 10;
int n, m, t, num[MAXN], h[MAXN], d[MAXN], in[MAXN], out[MAXN];
int o[4 * MAXN], p[4 * MAXN];
int lo[4 * MAXN], lp[4 * MAXN];
vector<int> graph[MAXN];
void dfs(int u, int f) {
  d[++t] = u;
  in[u] = t;
  out[u] = t;
  h[u] = h[f] + 1;
  for (auto v : graph[u]) {
    if (v != f) {
      dfs(v, u);
      out[u] = max(out[u], out[v]);
    }
  }
}
void build(int u, int l, int r) {
  if (l == r) {
    if (h[d[l]] % 2 == 1) {
      o[u] = num[d[l]];
    } else {
      p[u] = num[d[l]];
    }
    return;
  }
  int mid = (l + r) / 2;
  build(2 * u, l, mid);
  build(2 * u + 1, mid + 1, r);
  o[u] = o[2 * u] + o[2 * u + 1];
  p[u] = p[2 * u] + p[2 * u + 1];
}
void upd(int u, int l, int r, int a, int b, int val) {
  if (lo[u] != 0) {
    if (l != r) {
      lo[2 * u] += lo[u];
      lo[2 * u + 1] += lo[u];
    }
    o[u] += lo[u] * (r - l + 1);
    lo[u] = 0;
  }
  if (lp[u] != 0) {
    if (l != r) {
      lp[2 * u] += lp[u];
      lp[2 * u + 1] += lp[u];
    }
    p[u] += lp[u] * (r - l + 1);
    lp[u] = 0;
  }
  if (l > b || r < a) return;
  if (a <= l && r <= b) {
    if (r != l) {
      lo[2 * u] += val;
      lo[2 * u + 1] += val;
      lp[2 * u] += (-val);
      lp[2 * u + 1] += (-val);
    }
    o[u] += val * (r - l + 1);
    p[u] += (-val) * (r - l + 1);
    return;
  }
  int mid = (l + r) / 2;
  upd(2 * u, l, mid, a, b, val);
  upd(2 * u + 1, mid + 1, r, a, b, val);
  o[u] = o[2 * u] + o[2 * u + 1];
  p[u] = p[2 * u] + p[2 * u + 1];
}
int get(int u, int l, int r, int i) {
  if (lo[u] != 0) {
    if (l != r) {
      lo[2 * u] += lo[u];
      lo[2 * u + 1] += lo[u];
    }
    o[u] += lo[u] * (r - l + 1);
    lo[u] = 0;
  }
  if (lp[u] != 0) {
    if (l != r) {
      lp[2 * u] += lp[u];
      lp[2 * u + 1] += lp[u];
    }
    p[u] += lp[u] * (r - l + 1);
    lp[u] = 0;
  }
  if (l == r) {
    if (h[d[l]] % 2 == 1) {
      return o[u];
    } else {
      return p[u];
    }
  }
  int mid = (l + r) / 2;
  if (i <= mid)
    return get(2 * u, l, mid, i);
  else
    return get(2 * u + 1, mid + 1, r, i);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &num[i]);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(1, 1);
  build(1, 1, t);
  for (int i = 0; i < m; ++i) {
    int id;
    scanf("%d", &id);
    if (id == 1) {
      int x, val;
      scanf("%d%d", &x, &val);
      if (h[x] % 2 == 0) val = -val;
      upd(1, 1, t, in[x], out[x], val);
    } else {
      int x;
      scanf("%d", &x);
      printf("%d\n", get(1, 1, t, in[x]));
    }
  }
  return 0;
}
