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;
struct elist {
  int v, last;
} e[400010];
int n, m, a[200010], zh[200010], etot, L[200010], R[200010], tot, tt[200010];
int flag[200010];
inline void eadd(int u, int v) {
  e[++etot].v = v;
  e[etot].last = zh[u];
  zh[u] = etot;
  e[++etot].v = u;
  e[etot].last = zh[v];
  zh[v] = etot;
}
inline int lowbit(int x) { return (x & (-x)); }
inline void Init() {
  int u, v;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &u, &v);
    eadd(u, v);
  }
}
void dfs(int p, int x, int step) {
  flag[x] = step;
  L[x] = ++tot;
  for (int i = zh[x]; i; i = e[i].last)
    if (e[i].v != p) {
      dfs(x, e[i].v, -step);
    }
  R[x] = tot;
}
inline void Update(int x, int y) {
  for (int p = x; p <= n; p += lowbit(p)) tt[p] += y;
}
inline int Get_Sum(int x) {
  int tot = 0;
  for (int p = L[x]; p; p -= lowbit(p)) tot += tt[p];
  return tot;
}
inline void solve() {
  int x, y, z;
  Init();
  dfs(0, 1, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    if (x == 1) {
      scanf("%d", &z);
      z *= flag[y];
      Update(L[y], z);
      Update(R[y] + 1, -z);
    } else {
      printf("%d\n", a[y] + flag[y] * Get_Sum(y));
    }
  }
}
int main() {
  solve();
  return 0;
}
