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 int long long
#define mod 1000000007
#define inv2 500000004
int n,m,a[3010],f[3010][3010],g[3010][3010],A,B,ans;
signed main(){
	scanf("%lld%lld",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%lld",&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<=m;i++){
		scanf("%lld%lld",&A,&B);
		for(int j=1;j<=n;j++){
			g[A][j]=(f[A][j]+f[B][j])%mod*inv2%mod;
			g[j][A]=(f[j][A]+f[j][B])%mod*inv2%mod;
			g[B][j]=g[A][j];
			g[j][B]=g[j][A];		
		}
		g[A][B]=g[B][A]=(f[A][B]+f[B][A])%mod*inv2%mod;
		for(int j=1;j<=n;j++){
			f[A][j]=g[A][j];
			f[B][j]=g[B][j];
			f[j][A]=g[j][A];
			f[j][B]=g[j][B];			
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			(ans+=f[j][i])%=mod;
	for(int i=1;i<=m;i++)
		(ans*=2)%=mod;
	printf("%lld\n",ans);
}