CodeForces - 1093D Beautiful Graph 匹配问题

You are given an undirected unweighted graph consisting of n vertices and m

edges.

You have to write a number on each vertex of the graph. Each number should be 1

, 2 or 3

. The graph becomes beautiful if for each edge the sum of numbers on vertices connected by this edge is odd.

Calculate the number of possible ways to write numbers 1

, 2 and 3 on vertices so the graph becomes beautiful. Since this number may be large, print it modulo 998244353

Note that you have to write exactly one number on each vertex.

The graph does not have any self-loops or multiple edges.

Input

The first line contains one integer t

(1≤t≤3⋅105

) — the number of tests in the input.

The first line of each test contains two integers n

and m (1≤n≤3⋅105,0≤m≤3⋅105) — the number of vertices and the number of edges, respectively. Next m lines describe edges: i-th line contains two integers ui, vi (1≤ui,vi≤n;ui≠vi) — indices of vertices connected by i

-th edge.

It is guaranteed that ∑i=1tn≤3⋅105

and ∑i=1tm≤3⋅105

Output

For each test print one line, containing one integer — the number of possible ways to write numbers 1

, 2, 3 on the vertices of given graph so it becomes beautiful. Since answers may be large, print them modulo 998244353

Example

Input

Output

4
0

Note

Possible ways to distribute numbers in the first test:

the vertex 1

should contain 1, and 2 should contain 2

;
the vertex 1
should contain 3, and 2 should contain 2
;
the vertex 1
should contain 2, and 2 should contain 1
;
the vertex 1
should contain 2, and 2 should contain 3
1. .
In the second test there is no way to distribute numbers.

题意：1 2 3 ，每个点给一个值，要求相邻两点权值加和为奇数，求方案数

题解：假设有两个数1 2的话，很明显，如果存在合理方案，那么就肯定为两种，1和2换一下即可，因为这里多了一个3，所以，

如果权值为奇数的为cnt1个，偶数的cnt2个，此时方案数为2^cnt1, 然后奇偶一换，方案数为2^cnt2

所以方案总数为2^cnt1+2^cnt2,因为没说是所有的点联通，所以每块乘起来即可，若遇到不符合标记一下即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const ll mod=998244353; 
const int N=3e5+10;
#define pb push_back
int n,m,vis[N];
vector<int> v[N];
int cnt1,cnt2;
int flag;
void dfs(int u)
{
	if(vis[u]==1) cnt1++;
	else cnt2++;
	for(int i=0;i<v[u].size();i++)
	{
		int to=v[u][i];
		if(vis[to])
		{
			if(vis[to]==vis[u]) flag=1;  //   遇见不符合 标记
		}
		else
		{
			vis[to]=3-vis[u];
			dfs(to);
		}
	}
}
ll ksm(ll a,ll b)
{
	ll ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int x,y;
		scanf("%d%d",&n,&m);
		for(int i=0;i<=n;i++)vis[i]=0,v[i].clear();
		flag=0;
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&x,&y);
			v[x].pb(y);
			v[y].pb(x); 
		}
		ll ans=1;
		for(int i=1;i<=n;i++)
		{
			if(vis[i]) continue;
			vis[i]=1;
			cnt1=cnt2=0;
			dfs(i);
			if(flag) break;
			ans=(ans*(ksm(2,cnt1)+ksm(2,cnt2))%mod)%mod;
		}
		if(flag) cout<<"0\n";
		else cout<<ans<<endl;
	}
	return 0;
}

CodeForces - 1093D Beautiful Graph 匹配问题

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