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<bits/stdc++.h>
using namespace std;
#define int long long
const int N=3005,M=1e9+7;
int f[N][N],a[N],g[N][N],n,Q,x,y;
int ksm(int x,int y){
	if (!y)return 1;
	int z=ksm(x,y/2);
	z*=z;z%=M;
	if (y&1)z*=x;
	return z%M;
}
signed main(){
	scanf("%lld%lld",&n,&Q);
	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++)
			if (a[i]>a[j])f[i][j]=1;
	int K=ksm(2,Q),Inv=ksm(2,M-2);		
	while (Q--){
		scanf("%lld%lld",&x,&y);
		g[x][y]=f[y][x];
		g[y][x]=f[x][y];
		f[y][x]=(f[y][x]*Inv+g[y][x]*Inv)%M;
		f[x][y]=(f[x][y]*Inv+g[x][y]*Inv)%M;
		for (int i=1;i<=n;i++)
			if (i!=x&&i!=y)g[x][i]=f[y][i],g[i][x]=f[i][y];
		for (int i=1;i<=n;i++)
			if (i!=x&&i!=y)g[y][i]=f[x][i],g[i][y]=f[i][x];
		for (int i=1;i<=n;i++)
			if (i!=x&&i!=y){
				f[x][i]=(g[x][i]*Inv+f[x][i]*Inv)%M;
				f[y][i]=(g[y][i]*Inv+f[y][i]*Inv)%M;
				f[i][x]=(g[i][x]*Inv+f[i][x]*Inv)%M;
				f[i][y]=(g[i][y]*Inv+f[i][y]*Inv)%M;
			}
		for (int i=1;i<=n;i++)g[i][x]=g[i][y]=g[x][i]=g[y][i]=0;
	}
	int ans=0;
	for (int i=1;i<n;i++)
		for (int j=i+1;j<=n;j++)(ans+=f[i][j])%=M;
	printf("%lld\n",ans*K%M);	
}