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;
int n, m, cnt, w[400050], a[400050], c[2][400050];
int dp[400050], rk[400050], kth[400050], size[400050];
int tot, last[400050];
struct Edge {
  int from, to, s;
} edges[400050 << 1];
template <typename T>
void read(T &x) {
  long long k = 0;
  char c = getchar();
  x = 0;
  while (!isdigit(c) && c != EOF) k ^= c == '-', c = getchar();
  if (c == EOF) exit(0);
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x = k ? -x : x;
}
void read_char(char &c) {
  while (!isalpha(c = getchar()) && c != EOF)
    ;
}
void AddEdge(int x, int y) {
  edges[++tot] = Edge{x, y, last[x]};
  last[x] = tot;
}
void dfs(int x, int pre) {
  dp[x] = 1 - dp[pre];
  rk[x] = ++cnt;
  kth[cnt] = x;
  size[x] = 1;
  for (int i = last[x]; i; i = edges[i].s) {
    Edge &e = edges[i];
    if (e.to == pre) continue;
    dfs(e.to, x);
    size[x] += size[e.to];
  }
}
void update(int *c, int x, int tt) {
  while (x <= n) {
    c[x] += tt;
    x += x & -x;
  }
}
int get_sum(int *c, int x) {
  int ans = 0;
  while (x) {
    ans += c[x];
    x -= x & -x;
  }
  return ans;
}
int main() {
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) read(w[i]);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    read(x);
    read(y);
    AddEdge(x, y);
    AddEdge(y, x);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; i++) {
    int id, x, val;
    read(id);
    if (id == 1) {
      read(x);
      read(val);
      update(c[dp[x]], rk[x], val);
      update(c[dp[x]], rk[x] + size[x], -val);
    }
    if (id == 2) {
      read(x);
      int ans = get_sum(c[dp[x]], rk[x]) - get_sum(c[1 - dp[x]], rk[x]);
      ans += w[x];
      printf("%d\n", ans);
    }
  }
}
