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<iostream>
#include<cstdio>
#define LL long long
using namespace std;
int n,q,ans,er=1;
const int N=3010,mod=1000000007,inv2=(mod+1)/2;
int a[N],pos1[N],pos2[N],dp[N][N],tmp[N][N];
int main()
{
	cin>>n>>q;
	for(int i=1;i<=n;++i)scanf("%d",&a[i]);
	for(int i=1;i<=q;++i)scanf("%d%d",&pos1[i],&pos2[i]);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			dp[i][j]=a[i]>a[j];
	for(int i=1;i<=q;++i)
	{
		for(int j=1;j<=n;++j)
			if(j!=pos1[i]&&j!=pos2[i])
			{
				tmp[j][pos1[i]]=tmp[j][pos2[i]]=(LL)inv2*(dp[j][pos1[i]]+dp[j][pos2[i]])%mod;
				tmp[pos1[i]][j]=tmp[pos2[i]][j]=(LL)inv2*(dp[pos1[i]][j]+dp[pos2[i]][j])%mod;
			}
		for(int j=1;j<=n;++j)
			if(j!=pos1[i]&&j!=pos2[i])
			{
				dp[j][pos1[i]]=dp[j][pos2[i]]=tmp[j][pos1[i]];
				dp[pos1[i]][j]=dp[pos2[i]][j]=tmp[pos1[i]][j];
			}
		dp[pos1[i]][pos2[i]]=dp[pos2[i]][pos1[i]]=(LL)inv2*(dp[pos1[i]][pos2[i]]+dp[pos2[i]][pos1[i]])%mod;
		er=(er<<1)%mod;
	}
	for(int i=1;i<=n;++i)
		for(int j=i+1;j<=n;++j)
			(ans+=dp[i][j])%=mod;
	cout<<(LL)ans*er%mod;;
	return 0;
}