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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <functional>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cassert>
#define SIZE 3005
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

ll dp[SIZE][SIZE];
ll nx[SIZE][SIZE];
int X[SIZE],Y[SIZE];
int A[SIZE];

int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	for(int i=0;i<n;i++) scanf("%d",&A[i]);
	for(int i=0;i<m;i++) scanf("%d %d",&X[i],&Y[i]),X[i]--,Y[i]--;
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) if(A[i]<A[j]) dp[i][j]=1;
	ll inv=(MOD+1)/2;
	for(int i=0;i<m;i++)
	{
		for(int j=0;j<n;j++)
		{
			if(j!=X[i]&&j!=Y[i])
			{
				ll s=(dp[j][X[i]]+dp[j][Y[i]])%MOD*inv%MOD;
				dp[j][X[i]]=dp[j][Y[i]]=s;
				s=(dp[X[i]][j]+dp[Y[i]][j])%MOD*inv%MOD;
				dp[X[i]][j]=dp[Y[i]][j]=s;
			}
		}
		ll s=(dp[Y[i]][X[i]]+dp[X[i]][Y[i]])%MOD*inv%MOD;
		dp[X[i]][Y[i]]=dp[Y[i]][X[i]]=s;
	}
	ll ret=0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<i;j++)
		{
			ret+=dp[i][j];
			if(ret>=MOD) ret-=MOD;
		}
	}
	for(int i=0;i<m;i++) ret=ret*2LL%MOD;
	printf("%lld\n",ret);
	return 0;
}