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<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define re register
#define ll long long
inline int gi()
{
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=3010,Mod=1e9+7,INV2=500000004;
int n,Q,a[N],f[N][N];

int main()
{
	n=gi();Q=gi();int pw=1;
	for(int i=1;i<=n;i++)a[i]=gi();
	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=gi(),y=gi();pw=2ll*pw%Mod;
		f[x][y]=f[y][x]=1ll*INV2*(f[x][y]+f[y][x])%Mod;
		for(int i=1;i<=n;i++)
			if(x!=i&&i!=y)
			{
				f[x][i]=f[y][i]=1ll*INV2*(f[x][i]+f[y][i])%Mod;
				f[i][x]=f[i][y]=1ll*INV2*(f[i][x]+f[i][y])%Mod;
			}
	}
	int ans=0;
	for(int i=1;i<=n;i++)
		for(int j=1;j<i;j++)ans=(ans+f[i][j])%Mod;
	printf("%lld\n",1ll*ans*pw%Mod);
	return 0;
}