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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
const int N = 2e5 + 5, M = 2 * N + 5;
int head[N], to[M], nxt[M], ne, n;
void init() {
  ne = 0;
  memset(head, -1, n * sizeof head[0]);
}
void addEdge(int f, int t) {
  to[ne] = t;
  nxt[ne] = head[f];
  head[f] = ne++;
}
void addBiEdge(int a, int b) {
  addEdge(a, b);
  addEdge(b, a);
}
int dep[N];
int in[N], out[N], id(-1);
int nodes[N], sz;
void dfs(int u = 0, int par = -1, int l = 0) {
  dep[u] = l;
  in[u] = ++id;
  nodes[sz++] = u;
  for (int e = head[u]; ~e; e = nxt[e]) {
    int v = to[e];
    if (v == par) continue;
    dfs(v, u, l + 1);
  }
  out[u] = id;
}
int a[N];
long long tree[1 << 19][2], lazy[1 << 19][2];
void pre() { dfs(); }
void pushDown(int ni, int ns, int ne, int id) {
  if (ns != ne) {
    lazy[ni * 2][id] += lazy[ni][id];
    lazy[ni * 2 + 1][id] += lazy[ni][id];
  } else if ((id & 1) == (dep[nodes[ns]] & 1))
    tree[ni][id] += lazy[ni][id];
  lazy[ni][id] = 0;
}
void update(int ni, int ns, int ne, int id, int qs, int qe, int val) {
  if (ns > qe || ne < qs) return;
  if (qs <= ns && ne <= qe) {
    lazy[ni][id] += val;
    return;
  }
  pushDown(ni, ns, ne, id);
  int mid = (ns + (ne - ns) / 2);
  int lf = ni * 2, rt = lf + 1;
  update(lf, ns, mid, id, qs, qe, val);
  update(rt, mid + 1, ne, id, qs, qe, val);
}
long long get(int ni, int ns, int ne, int id, int qi) {
  pushDown(ni, ns, ne, id);
  if (ns == ne) return tree[ni][id];
  int mid = (ns + (ne - ns) / 2);
  int lf = ni * 2, rt = lf + 1;
  if (mid >= qi) return get(lf, ns, mid, id, qi);
  return get(rt, mid + 1, ne, id, qi);
}
void tst(int ni, int ns, int ne) {
  pushDown(ni, ns, ne, 0);
  pushDown(ni, ns, ne, 1);
  if (ns == ne) {
    cout << "Idx = " << nodes[ns] + 1 << " --> " << tree[ni][0] << ", "
         << tree[ni][1] << endl;
    return;
  }
  int mid = (ns + (ne - ns) / 2);
  int lf = ni * 2, rt = lf + 1;
  tst(lf, ns, mid);
  tst(rt, mid + 1, ne);
}
int main() {
  int q;
  scanf("%d %d", &n, &q);
  init();
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    addBiEdge(--u, --v);
  }
  pre();
  while (q--) {
    int typ, u;
    scanf("%d %d", &typ, &u);
    --u;
    if (typ == 2) {
      printf("%lld\n",
             get(1, 0, n - 1, 0, in[u]) + get(1, 0, n - 1, 1, in[u]) + a[u]);
      continue;
    }
    int val;
    scanf("%d", &val);
    update(1, 0, n - 1, (dep[u] & 1), in[u], out[u], val);
    update(1, 0, n - 1, !(dep[u] & 1), in[u], out[u], -val);
  }
}
