1332_F. Independent Set
Eric is the teacher of graph theory class. Today, Eric teaches independent set and edge-induced subgraph.

Given a graph G=(V,E), an independent set is a subset of vertices V' ⊂ V such that for every pair u,v ∈ V', (u,v) not ∈ E (i.e. no edge in E connects two vertices from V').

An edge-induced subgraph consists of a subset of edges E' ⊂ E and all the vertices in the original graph that are incident on at least one edge in the subgraph.

Given E' ⊂ E, denote G[E'] the edge-induced subgraph such that E' is the edge set of the subgraph. Here is an illustration of those definitions:

<image>

In order to help his students get familiar with those definitions, he leaves the following problem as an exercise:

Given a tree G=(V,E), calculate the sum of w(H) over all except null edge-induced subgraph H of G, where w(H) is the number of independent sets in H. Formally, calculate ∑ _{∅ not= E' ⊂ E} w(G[E']).

Show Eric that you are smarter than his students by providing the correct answer as quickly as possible. Note that the answer might be large, you should output the answer modulo 998,244,353.

Input

The first line contains a single integer n (2 ≤ n ≤ 3 ⋅ 10^5), representing the number of vertices of the graph G.

Each of the following n-1 lines contains two integers u and v (1 ≤ u,v ≤ n, u not= v), describing edges of the given tree.

It is guaranteed that the given edges form a tree.

Output

Output one integer, representing the desired value modulo 998,244,353.

Examples

Input


2
2 1


Output


3


Input


3
1 2
3 2


Output


11

Note

For the second example, all independent sets are listed below.

<image>
```python
import sys
readline = sys.stdin.readline

def parorder(Edge, p):
    N = len(Edge)
    par = [0]*N
    par[p] = -1
    stack = [p]
    order = []
    visited = set([p])
    ast = stack.append
    apo = order.append
    while stack:
        vn = stack.pop()
        apo(vn)
        for vf in Edge[vn]:
            if vf in visited:
                continue
            visited.add(vf)
            par[vf] = vn
            ast(vf)
    return par, order

def getcld(p):
    res = [[] for _ in range(len(p))]
    for i, v in enumerate(p[1:], 1):
        res[v].append(i)
    return res


N = int(readline())
MOD = 998244353
Edge = [[] for _ in range(N)]
for _ in range(N-1):
    a, b = map(int, readline().split())
    a -= 1
    b -= 1
    Edge[a].append(b)
    Edge[b].append(a)

P, L = parorder(Edge, 0)
C = getcld(P)

dp = [[1, 1, 0, 0, 1] for _ in range(N)]
for p in L[::-1]:
    if not C[p]:
        continue
    res = 1
    res2 = 1
    res3 = 1
    for ci in C[p]:
        res = (res*(dp[ci][2] + dp[ci][3] + dp[ci][4])) % MOD
        res2 = (res2*(dp[ci][1] + dp[ci][2] + 2*dp[ci][3] + dp[ci][4])) % MOD
        res3 = (res3*(sum(dp[ci]) + dp[ci][2] + dp[ci][3])) % MOD
    dp[p][0] = res
    dp[p][1] = res
    dp[p][2] = (res2 - res)%MOD
    dp[p][3] = (res3 - res)%MOD
    dp[p][4] = res
print((dp[0][2] + dp[0][3] + dp[0][4] - 1) %MOD)
```