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;
template <class T>
inline T BIGMOD(T n, T m, T mod) {
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (ans > mod) ans %= mod;
    }
    k *= k;
    if (k > mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
template <class T>
inline T power(T b, T p) {
  if (p < 0) return -1;
  if (b <= 0) return -2;
  if (!p) return 1;
  return b * power(b, p - 1);
}
template <typename T>
inline bool isOn(T mask, T pos) {
  return mask & (1 << pos);
}
template <typename T>
inline int Off(T mask, T pos) {
  return mask ^ (1 << pos);
}
template <typename T>
inline int On(T mask, T pos) {
  return mask | (1 << pos);
}
long long X;
long long flag[2][1 << 19];
int in[2 * 1000005], node[2 * 1000005], l[2 * 1000005];
int range[2 * 1000005], a[2 * 1000005];
vector<int> adj[2 * 1000005];
int n, m, tn;
void dfs(int cn, int cp) {
  in[cn] = ++tn;
  for (int i = 0; i < adj[cn].size(); i++)
    if (adj[cn][i] != cp) {
      l[adj[cn][i]] = l[cn] ^ 1;
      dfs(adj[cn][i], cn);
      range[cn] += 1 + range[adj[cn][i]];
    }
}
void insert(int cs, int cst, int ce, int s, int e, int cn, int val) {
  if (cst == s && ce == e) {
    flag[l[cn]][cs] += val;
    return;
  }
  int m = (s + e) >> 1;
  for (int i = 0; i < 2; i++) {
    flag[i][2 * cs] += flag[i][cs];
    flag[i][2 * cs + 1] += flag[i][cs];
    flag[i][cs] = 0;
  }
  if (cst <= m) insert(2 * cs, cst, min(m, ce), s, m, cn, val);
  if (ce > m) insert(2 * cs + 1, max(cst, m + 1), ce, m + 1, e, cn, val);
}
int findNo(int cs, int in, int s, int e) {
  if (s == e) {
    int cn = node[in];
    int cl = l[cn];
    return flag[cl][cs] - flag[cl ^ 1][cs];
  }
  int m = (s + e) >> 1;
  for (int i = 0; i < 2; i++) {
    flag[i][2 * cs] += flag[i][cs];
    flag[i][2 * cs + 1] += flag[i][cs];
    flag[i][cs] = 0;
  }
  if (in <= m) return findNo(2 * cs, in, s, m);
  return findNo(2 * cs + 1, in, m + 1, e);
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  l[1] = 1;
  dfs(1, 1);
  for (int i = 1; i <= n; i++) node[in[i]] = i;
  for (int i = 1; i <= n; i++) insert(1, i, i, 1, n, node[i], a[node[i]]);
  for (int i = 0; i < m; i++) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int cn, val;
      scanf("%d%d", &cn, &val);
      insert(1, in[cn], in[cn] + range[cn], 1, n, cn, val);
    } else {
      int cn;
      scanf("%d", &cn);
      printf("%d\n", findNo(1, in[cn], 1, n));
    }
  }
  return 0;
}
