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 R register
#define ll long long
const int MAXN=3e3+10;
const int mod=1e9+7;

inline int qpow(int x,int p)
{
	int res=1;
	while(p)
	{
		if(p&1) res=(ll)res*x%mod;
		x=(ll)x*x%mod;
		p>>=1;
	}
	return res;
}

int n,q;
int a[MAXN];
int f[MAXN][MAXN];

int main()
{
	scanf("%d%d",&n,&q);
	for(R int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(R int i=1;i<=n;i++)
		for(R int j=1;j<=n;j++)
			f[i][j]=a[i]<a[j]?1:0;
	int inv=qpow(2,mod-2);
	for(R int i=1;i<=q;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		f[x][y]=f[y][x]=(ll)(f[x][y]+f[y][x])*inv%mod;
		for(R int j=1;j<=n;j++)
		{
			if(j==x||j==y) continue;
			f[j][x]=f[j][y]=(ll)(f[j][x]+f[j][y])*inv%mod;
			f[x][j]=f[y][j]=(ll)(f[x][j]+f[y][j])*inv%mod;
		}
	}
	int ans=0;
	for(R int i=1;i<=n;i++)
		for(R int j=1;j<i;j++)
			ans=(ans+f[i][j])>mod?ans+f[i][j]-mod:ans+f[i][j];
	printf("%lld\n",(ll)ans*qpow(2,q)%mod);
	return 0;
}