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 = 2e5 + 10;
int val[N];
vector<int> g[N];
int timeLabel;
int ll[N], rr[N], dep[N], sum[2][N << 2], lazy[2][N << 2];
void dfs(int u, int fa, int d) {
  ll[u] = ++timeLabel;
  dep[u] = d;
  int cnt = (int)g[u].size();
  for (int i = 0; i < cnt; ++i) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs(v, u, d + 1);
  }
  rr[u] = timeLabel;
}
void pushUp(int rt, int x) {
  sum[x][rt] = sum[x][rt << 1] + sum[x][rt << 1 | 1];
}
void build(int rt, int l, int r, int x) {
  sum[x][rt] = lazy[x][rt] = 0;
  if (l == r) {
    return;
  }
  int m = (l + r) >> 1;
  build(rt << 1, l, m, x);
  build(rt << 1 | 1, m + 1, r, x);
  pushUp(rt, x);
}
void pushDown(int rt, int len, int x) {
  if (lazy[x][rt]) {
    lazy[x][rt << 1] += lazy[x][rt];
    lazy[x][rt << 1 | 1] += lazy[x][rt];
    sum[x][rt << 1] += lazy[x][rt];
    sum[x][rt << 1 | 1] += lazy[x][rt];
    lazy[x][rt] = 0;
  }
}
void update(int rt, int l, int r, int L, int R, int data, int x) {
  if (L <= l && r <= R) {
    lazy[x][rt] += data;
    sum[x][rt] += data;
    return;
  }
  pushDown(rt, r - l + 1, x);
  int m = (l + r) >> 1;
  if (L <= m) {
    update(rt << 1, l, m, L, R, data, x);
  }
  if (R > m) {
    update(rt << 1 | 1, m + 1, r, L, R, data, x);
  }
  pushUp(rt, x);
}
int query(int rt, int l, int r, int L, int R, int x) {
  if (L <= l && r <= R) {
    return sum[x][rt];
  }
  pushDown(rt, r - l + 1, x);
  int m = (l + r) >> 1;
  int ret = 0;
  if (L <= m) {
    ret += query(rt << 1, l, m, L, R, x);
  }
  if (R > m) {
    ret += query(rt << 1 | 1, m + 1, r, L, R, x);
  }
  return ret;
}
int S() {
  int ret = 0, neg = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') neg = 1;
  for (; c >= '0' && c <= '9'; c = getchar())
    ret = (ret << 3) + (ret << 1) + c - '0';
  return neg ? -ret : ret;
}
int main() {
  int n = S(), m = S();
  for (int i = 1; i <= n; ++i) val[i] = S();
  for (int i = 1; i < n; ++i) {
    int u = S(), v = S();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1, 1);
  for (int i = 0; i < 2; ++i) {
    build(1, 1, n, i);
  }
  while (m--) {
    int op = S(), x = S(), data;
    if (op == 1) {
      data = S();
      update(1, 1, n, ll[x], rr[x], data, dep[x] & 1);
      if (ll[x] < rr[x]) {
        update(1, 1, n, ll[x] + 1, rr[x], -data, (dep[x] & 1) ^ 1);
      }
    } else {
      int ans = query(1, 1, n, ll[x], ll[x], dep[x] & 1) + val[x];
      printf("%d\n", ans);
    }
  }
  return 0;
}
