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:
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 3001
#define mdn 1000000007
#define inv 500000004ll 
using namespace std;

int f[N][N],a[N],n,q;
int ksm(int bs,int mi)
{
	int ans=1;
	while(mi)
	{
		if(mi&1)	ans=1ll*ans*bs%mdn;
		bs=1ll*bs*bs%mdn; mi>>=1;// printf("QAQ");
	}
	return ans;
}
int main()
{
	scanf("%d%d",&n,&q); int remq=q;
	for(int i=1;i<=n;i++)	scanf("%d",&a[i]);
	for(int i=1;i<=n;i++)	for(int j=1;j<=n;j++)
		f[i][j]=a[i]>a[j];
	while(q--)
	{
		int x,y; scanf("%d%d",&x,&y);
		f[x][y]=f[y][x]=inv*(f[x][y]+f[y][x])%mdn;
		for(int i=1;i<=n;i++)	if(i!=x&&i!=y)
			f[x][i]=f[y][i]=inv*(f[x][i]+f[y][i])%mdn,
			f[i][x]=f[i][y]=inv*(f[i][x]+f[i][y])%mdn;
	}
	int ans=0; 
	for(int i=1;i<=n;i++)	for(int j=1;j<i;j++)
		ans=(ans+f[j][i])%mdn;
	printf("%d\n",1ll*ans*ksm(2,remq)%mdn);
	return 0;
}