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 bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c<'0' | c> '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const long long N = 200005;
const long long MOD = 1e9 + 7;
long long cost[N], cost2[N];
vector<long long> v[N];
set<long long> even, odd;
long long vis[N];
long long ck[N];
long long subr[N][2];
long long c5;
long long tree[N * 4][2];
long long lazy[N * 4][2];
void dfs(long long x, long long h) {
  vis[x] = 1;
  subr[x][0] = ++c5;
  ck[c5] = h & 1;
  if (h & 1)
    odd.emplace(c5);
  else
    even.emplace(c5);
  for (long long y : v[x])
    if (!vis[y]) dfs(y, h + 1);
  subr[x][1] = c5;
}
void build(long long node, long long l, long long r) {
  if (l == r) {
    tree[node][0] = cost[l];
    tree[node][1] = cost[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(node << 1, l, mid);
  build(node << 1 | 1, mid + 1, r);
}
void update(long long node, long long l, long long r, long long a, long long b,
            long long val, long long c) {
  if (lazy[node][c]) {
    tree[node][c] += lazy[node][c];
    if (l != r) {
      lazy[node << 1][c] += lazy[node][c];
      lazy[node << 1 | 1][c] += lazy[node][c];
    }
    lazy[node][c] = 0;
  }
  long long mid = (l + r) >> 1;
  if (l > b || r < a) return;
  if (l >= a && r <= b) {
    tree[node][c] += val;
    if (l != r) {
      lazy[node << 1][c] += val;
      lazy[node << 1 | 1][c] += val;
    }
    return;
  }
  update(node << 1, l, mid, a, b, val, c);
  update(node << 1 | 1, mid + 1, r, a, b, val, c);
}
long long query(long long node, long long l, long long r, long long p,
                long long c) {
  if (lazy[node][c]) {
    tree[node][c] += lazy[node][c];
    if (l != r) {
      lazy[node << 1][c] += lazy[node][c];
      lazy[node << 1 | 1][c] += lazy[node][c];
    }
    lazy[node][c] = 0;
  }
  if (l > p || r < p) return 0;
  if (l >= p && r <= p) return tree[node][c];
  long long mid = (l + r) >> 1;
  return query(node << 1, l, mid, p, c) +
         query(node << 1 | 1, mid + 1, r, p, c);
}
int main() {
  long long n = ({
    long long a;
    read(a);
    a;
  }),
            Q = ({
              long long a;
              read(a);
              a;
            });
  for (int i = 1; i <= n; i++)
    cost2[i] = ({
      long long a;
      read(a);
      a;
    });
  for (int i = 1; i <= n - 1; i++) {
    long long x = ({
      long long a;
      read(a);
      a;
    }),
              y = ({
                long long a;
                read(a);
                a;
              });
    v[x].emplace_back(y);
    v[y].emplace_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) cost[subr[i][0]] = cost2[i];
  build(1, 1, n);
  while (Q--) {
    long long q = ({
      long long a;
      read(a);
      a;
    }),
              x = ({
                long long a;
                read(a);
                a;
              });
    if (q == 1) {
      long long val = ({
        long long a;
        read(a);
        a;
      });
      long long a = subr[x][0];
      long long b = subr[x][1];
      long long y = ck[a];
      if (y) {
        update(1, 1, n, a, b, val, 1);
        update(1, 1, n, a, b, -val, 0);
      } else {
        update(1, 1, n, a, b, val, 0);
        update(1, 1, n, a, b, -val, 1);
      }
    } else {
      long long a = subr[x][0];
      long long y = ck[a];
      printf("%lld", query(1, 1, n, a, y)), printf("\n");
    }
  }
}
