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;
typedef long long ll;
const int N=3005,p=1000000007,inv2=(p+1)/2,inf=0x3f3f3f3f;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int qpow(int x,int k){
	int t=1;
	for (;k;k>>=1){
		if (k&1) t=(ll)t*x%p;
		x=(ll)x*x%p;
	}
	return t;
}
int n,m,ans,a[N],f[N][N];
void check(int &x,int y){if ((x+=y)>=p) x-=p;}
int main(){
	n=read();m=read();
	for(int i=1;i<=n;i++) a[i]=read();
	for (int i=1;i<=n;i++)
	for (int j=1;j<=n;j++) f[i][j]=(a[i]>a[j]);
	for (int k=1;k<=m;k++){
		int x=read(),y=read();
		f[x][y]=f[y][x]=(ll)(f[x][y]+f[y][x])*inv2%p;
		for (int i=1;i<=n;i++)
		if ((i!=x)&&(i!=y)){
			f[x][i]=f[y][i]=(ll)(f[x][i]+f[y][i])*inv2%p;
			f[i][x]=f[i][y]=(ll)(f[i][x]+f[i][y])*inv2%p;
		}
	}
	ans=0;
	for (int i=1;i<=n;i++)
	for (int j=i+1;j<=n;j++) check(ans,f[i][j]);
	ans=(ll)ans*qpow(2,m)%p;
	printf("%d\n",ans);
	return 0;
}