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>
#include <algorithm>

int f[3001][3001],n,q,x[3001],y[3001],a[3001],Pow[3001],ans;
const int P=1000000007;
inline int mul(const int &a,const int &b){return 1ll*a*b%P;}
inline int add(int a,const int &b){a+=b;return(a>=P)?a-P:a;}
inline int sub(int a,const int &b){a-=b;return (a<0)?a+P:a;}
int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)scanf("%d",a+i);
	Pow[0]=1;
	for(int i=1;i<=q;i++)scanf("%d%d",x+i,y+i),(x[i]>y[i])?std::swap(x[i],y[i]):(void)0,Pow[i]=add(Pow[i-1],Pow[i-1]);
	for(int i=q;i;i--){
		for(int j=1;j<=n;j++)
			if(j<x[i])f[j][x[i]]=f[j][y[i]]=mul(add(f[j][y[i]],f[j][x[i]]),(P+1)>>1);
			else if(j>x[i]&&j<y[i]){
				int tem=mul(add(f[x[i]][j],sub(1,f[j][y[i]])),(P+1)>>1);
				f[j][y[i]]=mul(add(f[j][y[i]],sub(1,f[x[i]][j])),(P+1)>>1);
				f[x[i]][j]=tem;
			}
			else if(j>y[i])f[x[i]][j]=f[y[i]][j]=mul(add(f[y[i]][j],f[x[i]][j]),(P+1)>>1);
		f[x[i]][y[i]]=(P+1)>>1;
	}
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			if(a[i]<a[j])ans=add(ans,f[i][j]);
			else if(a[i]>a[j])ans=add(ans,sub(1,f[i][j]));
	printf("%d\n",mul(ans,Pow[q]));
}