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<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gt getchar()
#define ll long long
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
inline int in()
{
	int k=0;char ch=gt;
	while(ch<'-')ch=gt;
	while(ch>'-')k=k*10+ch-'0',ch=gt;
	return k;
}
const int N=3005,YL=1e9+7,inv2=5e8+4;
inline int MO(const int &x){return x>=YL?x-YL:x;}
inline int ksm(int a,int k){int r=1;while(k){if(k&1)r=1ll*r*a%YL;a=1ll*a*a%YL,k>>=1;}return r;}
int a[N],f[N][N],g[4][N];
int main()
{
	int n=in(),Q=in();int ans=ksm(2,Q);
	for(int i=1;i<=n;++i)a[i]=in();
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			f[i][j]=(a[i]<a[j]);
	while(Q--)
	{
		int x=in(),y=in();
		for(int i=1;i<=n;++i)g[0][i]=f[x][i],g[1][i]=f[y][i];
		for(int i=1;i<=n;++i)g[2][i]=f[i][x],g[3][i]=f[i][y];
		for(int i=1;i<=n;++i)f[y][i]=f[x][i]=1ll*inv2*MO(g[0][i]+g[1][i])%YL;
		for(int i=1;i<=n;++i)f[i][x]=f[i][y]=1ll*inv2*MO(g[2][i]+g[3][i])%YL;
		f[x][y]=f[y][x]=1ll*inv2*MO(g[0][y]+g[1][x])%YL;
	}
	int sum=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<i;++j)
			sum=MO(sum+f[i][j]);
	printf("%lld\n",1ll*ans*sum%YL);
	return 0;
}