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 CO const
#define IN inline
typedef long long int64;

template<class T> IN T read(){
	T x=0,w=1;char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-') w=-w;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*w;
}
template<class T> IN T read(T&x){
	return x=read<T>();
}

CO int P=1e9+7,i2=(P+1)/2;
IN int add(int a,int b){
	return (a+=b)>=P?a-P:a;
}
IN int mul(int a,int b){
	return (int64)a*b%P;
}
IN int fpow(int a,int b){
	int ans=1;
	for(;b;b>>=1,a=mul(a,a))
		if(b&1) ans=mul(ans,a);
	return ans;
}

CO int N=3e3+10;
int A[N],F[N][N];

int main(){
	int n=read<int>(),m=read<int>();
	for(int i=1;i<=n;++i) read(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){
		int x=read<int>(),y=read<int>();
		F[x][y]=F[y][x]=mul(add(F[x][y],F[y][x]),i2);
		for(int j=1;j<=n;++j)if(j!=x and j!=y){
			F[x][j]=F[y][j]=mul(add(F[x][j],F[y][j]),i2);
			F[j][x]=F[j][y]=mul(add(F[j][x],F[j][y]),i2);
		}
	}
	int ans=0;
	for(int i=1;i<=n;++i)for(int j=i+1;j<=n;++j)
		ans=add(ans,F[i][j]);
	ans=mul(ans,fpow(2,m));
	printf("%d\n",ans);
	return 0;
}