Problem: You are given an integer sequence of length N: A_1,A_2,...,A_N. Let us perform Q operations in order. The i-th operation is described by two integers X_i and Y_i. In this operation, we will choose one of the following two actions and perform it:

* Swap the values of A_{X_i} and A_{Y_i}
* Do nothing



There are 2^Q ways to perform these operations. Find the sum of the inversion numbers of the final sequences for all of these ways to perform operations, modulo 10^9+7.

Here, the inversion number of a sequence P_1,P_2,...,P_M is the number of pairs of integers (i,j) such that 1\leq i < j\leq M and P_i > P_j.

Constraints

* 1 \leq N \leq 3000
* 0 \leq Q \leq 3000
* 0 \leq A_i \leq 10^9(1\leq i\leq N)
* 1 \leq X_i,Y_i \leq N(1\leq i\leq Q)
* X_i\neq Y_i(1\leq i\leq Q)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N Q
A_1
:
A_N
X_1 Y_1
:
X_Q Y_Q


Output

Print the sum of the inversion numbers of the final sequences, modulo 10^9+7.

Examples

Input

3 2
1
2
3
1 2
1 3


Output

6


Input

5 3
3
2
3
1
4
1 5
2 3
4 2


Output

36


Input

9 5
3
1
4
1
5
9
2
6
5
3 5
8 9
7 9
3 2
3 8


Output

425

Solution (in CPP):

Solution:
#include<cstdio>
#define N 3010
#define mod 1000000007
#define inv 500000004
int f[N][N],a[N];
int qpow(int base,int n)
{
	if (!n) return 1;
	int ans=base%mod;--n;
	while(n)
	{
		if (n&1) ans=1ll*ans*base%mod;
		base=1ll*base*base%mod;
		n>>=1;
	}
	return ans;
}
inline int read(){
	int n=0;char a;bool z=false;
	while(a=getchar())
	{
		if (a>'9'||a<'0')
			if (z) break;
			else continue;
		if (!z) z=true;
		n=(n<<1)+(n<<3)+(a^48);
	}
	return n;
}
int main()
{
	int n,Q,x,y,ans=0;
	scanf("%d%d",&n,&Q);
	for (int i=1;i<=n;++i) scanf("%d",&a[i]);
	for (int i=1;i<=n;++i)
		for (int j=1;j<=n;++j)
			f[i][j]=a[i]>a[j];
	for (int i=1;i<=Q;++i)
	{
		scanf("%d%d",&x,&y);
		f[x][y]=f[y][x]=1ll*(f[x][y]+f[y][x])*inv%mod;
		for (int j=1;j<=n;++j)
			if (j^x&&j^y)
			{
				f[j][x]=f[j][y]=1ll*(f[j][x]+f[j][y])*inv%mod;
				f[x][j]=f[y][j]=1ll*(f[x][j]+f[y][j])*inv%mod;
			}
	}
	for (int i=1;i<=n;++i)
		for (int j=1;j<i;++j)
			ans=(ans+f[j][i])>=mod?ans+f[j][i]-mod:ans+f[j][i];
	printf("%lld",1ll*ans*qpow(2,Q)%mod);
	return 0;
}

//by qlwpc