Problem:
Score : 700 points

Problem Statement

We have a tree with N vertices. Vertex 1 is the root of the tree, and the parent of Vertex i (2 \leq i \leq N) is Vertex P_i.

To each vertex in the tree, Snuke will allocate a color, either black or white, and a non-negative integer weight.

Snuke has a favorite integer sequence, X_1, X_2, ..., X_N, so he wants to allocate colors and weights so that the following condition is satisfied for all v.

The total weight of the vertices with the same color as v among the vertices contained in the subtree whose root is v, is X_v.

Here, the subtree whose root is v is the tree consisting of Vertex v and all of its descendants.

Determine whether it is possible to allocate colors and weights in this way.

Constraints

1 \leq N \leq 1 000

1 \leq P_i \leq i - 1

0 \leq X_i \leq 5 000

Inputs

Input is given from Standard Input in the following format:

N
P_2 P_3 ... P_N
X_1 X_2 ... X_N

Outputs

If it is possible to allocate colors and weights to the vertices so that the condition is satisfied, print POSSIBLE; otherwise, print IMPOSSIBLE.

Sample Input 1

3
1 1
4 3 2

Sample Output 1

POSSIBLE

For example, the following allocation satisfies the condition:

Set the color of Vertex 1 to white and its weight to 2.

Set the color of Vertex 2 to black and its weight to 3.

Set the color of Vertex 3 to white and its weight to 2.

There are also other possible allocations.

Sample Input 2

3
1 2
1 2 3

Sample Output 2

IMPOSSIBLE

If the same color is allocated to Vertex 2 and Vertex 3, Vertex 2 cannot be allocated a non-negative weight.

If different colors are allocated to Vertex 2 and 3, no matter which color is allocated to Vertex 1, it cannot be allocated a non-negative weight.

Thus, there exists no allocation of colors and weights that satisfies the condition.

Sample Input 3

8
1 1 1 3 4 5 5
4 1 6 2 2 1 3 3

Sample Output 3

POSSIBLE

Sample Input 4

1

0

Sample Output 4

POSSIBLE

Solution in LuaJIT (2.0.4) (judge verdict: Accepted)

local n = io.read("*n")
local p = {0}
for i = 2, n do
  p[i] = io.read("*n")
end
local base = {}
local t = {}
local x = {}
for i = 1, n do
  base[i] = 0
  t[i] = {}
  x[i] = io.read("*n")
end
local valid = true
for i = n, 1, -1 do
  local tgt = x[i] - base[i]
  if tgt < 0 then
    valid = false
    break
  end
  -- get highest value y where dp[y] == true
  local dp = {}
  for j = 1, tgt do
    dp[j] = false
  end
  local sum = base[i]
  for j = 1, #t[i] do
    local tv = t[i][j]
    sum = sum + tv
    for k = tgt - tv, 1, -1 do
      if dp[k] then dp[k + tv] = true end
    end
    if tv <= tgt then dp[tv] = true end
  end
  local r = 0
  for j = tgt, 1, -1 do
    if dp[j] then r = j break end
  end
  if 1 < i then
    local v1 = x[i] -- base[i] + r <= x[i]. "bese[i] + r" became x[i].
    local v2 = sum - base[i] - r
    local parent = p[i]
    if v2 < v1 then v1, v2 = v2, v1 end
    base[parent] = base[parent] + v1
    if v1 < v2 then table.insert(t[parent], v2 - v1) end
  end
end
print(valid and "POSSIBLE" or "IMPOSSIBLE")