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;
vector<long long int> p[200005], q[200005], f(400005, 0), b(400005, 0),
    a(200005), d(400005), e(400005);
map<pair<long long int, long long int>, long long int> mp;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
long long int comb(long long int i, long long int j) {
  long long int k = f[i];
  long long int g = (f[j] * (f[i - j])) % 998244353;
  long long int h = bpow(g, 998244353 - 2);
  return (k * h) % 998244353;
}
long long int t = 0, n, m;
void dfs(long long int i, long long int pr) {
  t++;
  d[i] = t;
  for (auto u : p[i]) {
    if (u != pr) {
      b[u] = -b[i];
      dfs(u, i);
    }
  }
  t++;
  e[i] = t;
}
void add(long long int i, long long int val) {
  while (i <= 2 * n) {
    f[i] += val;
    i += i & -i;
  }
}
long long int cal(long long int i) {
  long long int s = 0;
  while (i > 0) {
    s += f[i];
    i -= i & -i;
  }
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long int u, v;
  for (long long int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    p[u].push_back(v);
    p[v].push_back(u);
  }
  b[1] = 1;
  dfs(1, 0);
  while (m--) {
    long long int x, y, z;
    cin >> x;
    if (x == 1) {
      cin >> y >> z;
      add(d[y], b[y] * z);
      add(e[y] + 1, -b[y] * z);
    }
    if (x == 2) {
      cin >> y;
      long long int s = a[y] + b[y] * cal(e[y]);
      cout << s << "\n";
    }
  }
}
