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 = int(2e5) + 5;
bool type[N];
short int a[N];
int evn_t = 0;
int odd_t = 0;
vector<int> ver[N];
vector<int> evn_euler;
vector<int> odd_euler;
int seg0[(N << 1) + 100];
int seg1[(N << 1) + 100];
int odd_pos[N];
int evn_pos[N];
int odd_lt[N];
int odd_rt[N];
int evn_lt[N];
int evn_rt[N];
map<int, bool> isleaf0;
map<int, bool> isleaf1;
void dfs(int v, int p, int s) {
  if (s) {
    evn_lt[v] = evn_t++;
    evn_euler.push_back(v);
    evn_pos[v] = evn_euler.size() - 1;
    odd_lt[v] = odd_t;
  } else {
    odd_lt[v] = odd_t++;
    odd_euler.push_back(v);
    odd_pos[v] = odd_euler.size() - 1;
    type[v] = 1;
    evn_lt[v] = evn_t;
  }
  for (int i = 0; i < ver[v].size(); i++) {
    if (ver[v][i] != p) dfs(ver[v][i], v, 1 - s);
  }
  if (s) {
    evn_rt[v] = evn_t - 1;
    if (odd_lt[v] == odd_t)
      odd_lt[v] = odd_rt[v] = -1;
    else
      odd_rt[v] = odd_t - 1;
  } else {
    odd_rt[v] = odd_t - 1;
    if (evn_lt[v] == evn_t)
      evn_lt[v] = evn_rt[v] = -1;
    else
      evn_rt[v] = evn_t - 1;
  }
}
void build0(int l, int r, int id) {
  int m = (l + r) >> 1;
  if (l == r) {
    isleaf0[id] = 1;
    seg0[id] = 0;
    return;
  }
  build0(l, m, id << 1);
  build0(m + 1, r, id << 1 | 1);
}
void build1(int l, int r, int id) {
  int m = (l + r) >> 1;
  if (l == r) {
    isleaf1[id] = 1;
    seg1[id] = 0;
    return;
  }
  build1(l, m, id << 1);
  build1(m + 1, r, id << 1 | 1);
}
void pushf0(int id) {
  if (isleaf0[id]) {
  } else {
    seg0[id << 1] += seg0[id];
    seg0[id << 1 | 1] += seg0[id];
    seg0[id] = 0;
  }
}
void pushf1(int id) {
  if (isleaf1[id]) {
  } else {
    seg1[id << 1] += seg1[id];
    seg1[id << 1 | 1] += seg1[id];
    seg1[id] = 0;
  }
}
void q1(int val, int l, int r, int L, int R, int id) {
  int m = (L + R) >> 1;
  if (seg0[id] != 0) pushf0(id);
  if (r < L || R < l) return;
  if (l <= L && R <= r) {
    seg0[id] += val;
    return;
  }
  q1(val, l, r, L, m, id << 1);
  q1(val, l, r, m + 1, R, id << 1 | 1);
}
void q2(int val, int l, int r, int L, int R, int id) {
  int m = (L + R) >> 1;
  if (seg1[id] != 0) pushf1(id);
  if (r < L || R < l) return;
  if (l <= L && R <= r) {
    seg1[id] += val;
    return;
  }
  q2(val, l, r, L, m, id << 1);
  q2(val, l, r, m + 1, R, id << 1 | 1);
}
int find1(int pos, int L, int R, int id) {
  int m = (L + R) >> 1;
  if (seg0[id] != 0) pushf0(id);
  if (pos < L || pos > R) return 0;
  if (pos == L && pos == R) return seg0[id];
  int ans = find1(pos, L, m, id << 1) + find1(pos, m + 1, R, id << 1 | 1);
  return ans;
}
int find2(int pos, int L, int R, int id) {
  int m = (L + R) >> 1;
  if (seg1[id] != 0) pushf1(id);
  if (pos < L || pos > R) return 0;
  if (pos == L && pos == R) return seg1[id];
  int ans = find2(pos, L, m, id << 1) + find2(pos, m + 1, R, id << 1 | 1);
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    odd_lt[i] = -1;
    evn_lt[i] = -1;
    odd_rt[i] = -1;
    evn_rt[i] = -1;
  }
  for (int i = 0; i < n; i++) scanf("%hd", &a[i]);
  for (int i = 0; i < n - 1; i++) {
    int t1, t2;
    scanf("%d%d", &t1, &t2);
    ver[t1].push_back(t2);
    ver[t2].push_back(t1);
  }
  dfs(1, 0, 1);
  for (int i = 1; i <= n; i++) ver[i].clear();
  build0(0, evn_euler.size() - 1, 1);
  if (odd_euler.size() > 0) build1(0, odd_euler.size() - 1, 1);
  while (m--) {
    int q;
    scanf("%d", &q);
    if (q == 1) {
      int x, val;
      scanf("%d%d", &x, &val);
      if (evn_lt[x] != -1) {
        if (type[x] == 0)
          q1(val, evn_lt[x], evn_rt[x], 0, evn_euler.size() - 1, 1);
        else
          q1(-val, evn_lt[x], evn_rt[x], 0, evn_euler.size() - 1, 1);
      }
      if (odd_lt[x] != -1) {
        if (type[x] == 0)
          q2(-val, odd_lt[x], odd_rt[x], 0, odd_euler.size() - 1, 1);
        else
          q2(val, odd_lt[x], odd_rt[x], 0, odd_euler.size() - 1, 1);
      }
    } else {
      int x;
      scanf("%d", &x);
      int ans;
      if (type[x] == 0)
        ans = find1(evn_pos[x], 0, evn_euler.size() - 1, 1);
      else
        ans = find2(odd_pos[x], 0, odd_euler.size() - 1, 1);
      printf("%d\n", (int)a[x - 1] + ans);
    }
  }
  return 0;
}
